Formulas · Mathematics

Algebraic identities

11 formulas, each with worked examples. Revision sheet · Practise these formulas

Class 9

Square of a sum

(a + b)^{2} = a^{2} + 2 ab + b^{2}

What each letter means
$a$	the first term (a number or an expression)
$b$	the second term

Squaring a sum gives three parts, the square of each term plus twice their product, not just the two squares.

Why it works

(a + b)² means (a + b)(a + b). Multiplying out gives a² + ab + ba + b², and the two middle terms are equal, so they add to 2ab. As a picture, a square of side (a + b) splits into a square a², a square b², and two rectangles of area ab.

When to use it

Expanding a squared bracket, recognising a perfect square to factorise it, or squaring a number quickly by splitting it (103 = 100 + 3).

How to use it

Square the first term.
Add twice the product of the two terms.
Add the square of the second term.

To remember: First squared, plus twice first times last, plus last squared.

Other forms

Same formula, another form $(a + b)^{2} = (a + b) (a + b)$ the definition of squaring
Special case $(2 a)^{2} = 4 a^{2}$ When a = b.
Rearranged $a^{2} + b^{2} = (a + b)^{2} - 2 ab$ find a² + b² from a sum and a product
To remember itPicture a square of side (a + b) cut into a², b² and two ab rectangles.

Worked examples

Story

A square garden with 20 m sides is made 3 m longer on every side. What is the new area?

New side: 20 + 3 = 23 m
$(20 + 3)^{2} = 400 + 120 + 9$
$529 m^{2}$

Answer: 529

Picture

A square has side (a + b). Show its area as four pieces.

Cut the side into a and b on each edge.
The pieces are an a-by-a square, a b-by-b square and two a-by-b rectangles.
Total area: a² + ab + ab + b² = a² + 2ab + b².

Answer: a² + 2ab + b²

Expand

Expand (3x + 2)².

$(3 x)^{2} + 2 (3 x) (2) + 2^{2}$
$9 x^{2} + 12 x + 4$

Answer: 9x² + 12x + 4

Factorise

Factorise x² + 10x + 25.

Try it first, then show the working

$x^{2} = (x)^{2} and 25 = 5^{2}$
$10 x = 2 (x) (5), so it is a perfect square$
$x^{2} + 10 x + 25 = (x + 5)^{2}$

Answer: (x + 5)²

Simplify

Simplify (a + b)² − (a − b)².

Try it first, then show the working

$(a^{2} + 2 ab + b^{2}) - (a^{2} - 2 ab + b^{2})$
$a^{2} - a^{2} + 2 ab + 2 ab + b^{2} - b^{2}$
$4 ab$

Answer: 4ab

Prove

Show that (2a + 3b)² − 12ab = 4a² + 9b².

Try it first, then show the working

$(2 a + 3 b)^{2} = 4 a^{2} + 12 ab + 9 b^{2}$
$4 a^{2} + 12 ab + 9 b^{2} - 12 ab = 4 a^{2} + 9 b^{2}$

Answer: Both sides equal 4a² + 9b²

Quick calculation

Find 103² without a calculator.

Try it first, then show the working

$10 3^{2} = (100 + 3)^{2}$
$= 10 0^{2} + 2 (100) (3) + 3^{2}$
$= 10000 + 600 + 9 = 10609$

Answer: 10609

Common mistake: Writing (a + b)² = a² + b² and forgetting the middle term 2ab. Test with numbers: (3 + 4)² = 49, but 3² + 4² = 25.

Class 9

Square of a difference

(a - b)^{2} = a^{2} - 2 ab + b^{2}

What each letter means
$a$	the first term (a number or an expression)
$b$	the term being taken away

Squaring a difference gives the two squares minus twice the product. Only the middle term changes sign; the last term stays plus.

Why it works

Put −b in place of b in (a + b)² = a² + 2ab + b². The middle term becomes 2a(−b) = −2ab, and (−b)² = b², because a negative times a negative is positive.

When to use it

Expanding a squared bracket with a minus, spotting a perfect square with a minus middle term, or squaring a number just below a round one (98 = 100 − 2).

How to use it

Square the first term.
Subtract twice the product of the two terms.
Add the square of the second term.

To remember: Like the square of a sum, with a minus in the middle only.

Other forms

Same formula, another form $(a - b)^{2} = (b - a)^{2}$ Swapping the order does not change the square.
Same formula, another form $(a - b)^{2} = (a + b)^{2} - 4 ab$ Links the two squares: useful when you know a + b and ab.
Rearranged $a^{2} + b^{2} = (a - b)^{2} + 2 ab$ Find a² + b² from a difference and a product.

Worked examples

Expand

Expand (2x − 5)².

$(2 x - 5)^{2} = (2 x)^{2} - 2 (2 x) (5) + 5^{2}$
$= 4 x^{2} - 20 x + 25$

Answer: 4x² − 20x + 25

Factorise

Factorise 9y² − 12y + 4.

9y² is (3y)² and 4 is 2²; the middle term 12y is 2 × 3y × 2, with a minus.
$9 y^{2} - 12 y + 4 = (3 y - 2)^{2}$

Answer: (3y − 2)²

Simplify

Simplify (x − 3)² − (x − 5)².

$(x - 3)^{2} - (x - 5)^{2} = (x^{2} - 6 x + 9) - (x^{2} - 10 x + 25)$
$= 4 x - 16$

Answer: 4x − 16

Prove

Show that (a − b)² + 4ab = (a + b)².

Try it first, then show the working

$(a - b)^{2} + 4 ab = a^{2} - 2 ab + b^{2} + 4 ab$
$= a^{2} + 2 ab + b^{2} = (a + b)^{2}$

Answer: Both sides equal a² + 2ab + b²

Quick calculation

Find 98² without a calculator.

Try it first, then show the working

$9 8^{2} = (100 - 2)^{2}$
$= 10 0^{2} - 2 (100) (2) + 2^{2}$
$= 10000 - 400 + 4 = 9604$

Answer: 9604

Common mistake: Writing (a − b)² = a² − b², or giving the last term a minus sign. Test with numbers: (5 − 2)² = 9, and 25 − 20 + 4 = 9, but 25 − 4 = 21.

Class 9

Square of a sum of three terms

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 ab + 2 b c + 2 c a

What each letter means
$a$	the first term
$b$	the second term
$c$	the third term

Squaring three terms gives each term squared, plus twice the product of every pair.

Why it works

Treat (a + b) as one term and use the square of a sum: ((a + b) + c)² = (a + b)² + 2(a + b)c + c² = a² + 2ab + b² + 2ac + 2bc + c². As a picture, a square of side a + b + c splits into 3 squares and 6 rectangles, two of each pair.

When to use it

Expanding a bracket with three terms squared, finding a² + b² + c² from a + b + c and ab + bc + ca, or squaring numbers like 111 = 100 + 10 + 1.

How to use it

Square each of the three terms.
Multiply each pair (ab, bc, ca) and double it.
Add all six results, keeping each sign.

To remember: Every term squared, every pair doubled.

Other forms

Rearranged $a^{2} + b^{2} + c^{2} = (a + b + c)^{2} - 2 (ab + b c + c a)$ Find the sum of the squares from the sum and the sum of the pair products.
Same formula, another form $(a + b + c)^{2} = (a + b)^{2} + 2 (a + b) c + c^{2}$ Treat a + b as one term.

Worked examples

Expand

Expand (x + 2y + 3)².

$(x + 2 y + 3)^{2} = x^{2} + (2 y)^{2} + 3^{2} + 2 (x) (2 y) + 2 (2 y) (3) + 2 (3) (x)$
$= x^{2} + 4 y^{2} + 9 + 4 x y + 12 y + 6 x$

Answer: x² + 4y² + 9 + 4xy + 12y + 6x

Factorise

Factorise 4x² + y² + z² + 4xy + 2yz + 4zx.

The squares are (2x)², y² and z²; the pairs are 2(2x)(y) = 4xy, 2(y)(z) = 2yz and 2(z)(2x) = 4zx, all plus.
$4 x^{2} + y^{2} + z^{2} + 4 x y + 2 y z + 4 z x = (2 x + y + z)^{2}$

Answer: (2x + y + z)²

Simplify

Simplify (a + b + c)² − (a + b)².

$(a + b + c)^{2} - (a + b)^{2} = (a^{2} + b^{2} + c^{2} + 2 ab + 2 b c + 2 c a) - (a^{2} + 2 ab + b^{2})$
$= c^{2} + 2 b c + 2 c a$

Answer: c² + 2bc + 2ca

Prove

If a + b + c = 9 and ab + bc + ca = 26, find a² + b² + c².

Try it first, then show the working

a² + b² + c² = (a + b + c)² − 2(ab + bc + ca)
$9^{2} - 2 \times 26 = 81 - 52 = 29$

Answer: 29

Quick calculation

Find 111² by splitting 111 into 100 + 10 + 1.

Try it first, then show the working

$11 1^{2} = (100 + 10 + 1)^{2}$
$= 10000 + 100 + 1 + 2000 + 20 + 200$
$= 12321$

Answer: 12321

Common mistake: Missing one of the three pairs, or forgetting to double them. There are always 3 squares and 3 doubled pairs.

Class 9

Difference of two squares

(a + b) (a - b) = a^{2} - b^{2}

What each letter means
$a$	the first term
$b$	the second term

The sum of two terms times their difference is the first squared minus the second squared. Read backwards, any difference of two squares factorises.

Why it works

Multiply out: (a + b)(a − b) = a² − ab + ba − b². The middle terms −ab and +ab cancel, leaving a² − b². As a picture, cut a b-by-b square from the corner of an a-by-a square; the L shape left, of area a² − b², rearranges into an (a + b)-by-(a − b) rectangle.

When to use it

Factorising anything of the form (square) − (square), multiplying a sum by the matching difference, or multiplying numbers equally far either side of a round number (103 × 97).

How to use it

Check that both terms are squares and that there is a minus between them.
Find what is squared in each (a and b).
Write (a + b)(a − b).

To remember: Sum times difference gives difference of squares.

Other forms

Same formula, another form $a^{2} - b^{2} = (a - b) (a + b)$ The order of the two brackets does not matter.
Special case $a^{2} - 1 = (a + 1) (a - 1)$ When b = 1.

Worked examples

Picture

From a square of side 10 cm, a square of side 4 cm is cut from one corner. Find the area left, as a rectangle.

The L shape left rearranges into a rectangle (10 + 4) cm long and (10 − 4) cm wide.
$1 0^{2} - 4^{2} = (10 + 4) (10 - 4) = 14 \times 6 = 84$

Answer: 84

Expand

Expand (3x + 4)(3x − 4).

$(3 x + 4) (3 x - 4) = (3 x)^{2} - 4^{2}$
$= 9 x^{2} - 16$

Answer: 9x² − 16

Factorise

Factorise 49m² − 25n².

49m² is (7m)² and 25n² is (5n)².
$49 m^{2} - 25 n^{2} = (7 m + 5 n) (7 m - 5 n)$

Answer: (7m + 5n)(7m − 5n)

Simplify

Simplify (x + 1)(x − 1)(x² + 1).

Try it first, then show the working

$(x + 1) (x - 1) = x^{2} - 1$
$(x^{2} - 1) (x^{2} + 1) = x^{4} - 1$

Answer: x⁴ − 1

Prove

Show that x⁴ − y⁴ = (x − y)(x + y)(x² + y²), so x − y is a factor of x⁴ − y⁴.

Try it first, then show the working

$x^{4} - y^{4} = (x^{2})^{2} - (y^{2})^{2}$
$= (x^{2} - y^{2}) (x^{2} + y^{2})$
$= (x - y) (x + y) (x^{2} + y^{2})$

Answer: x⁴ − y⁴ = (x − y)(x + y)(x² + y²)

Quick calculation

Find 103 × 97 without a calculator.

Try it first, then show the working

$103 \times 97 = (100 + 3) (100 - 3)$
$= 10 0^{2} - 3^{2} = 10000 - 9 = 9991$

Answer: 9991

Common mistake: Trying to factorise a sum of two squares, a² + b², this way: it has no such factors. Also, (a − b)² is not a² − b².

Class 9

Product of (x + a) and (x + b)

(x + a) (x + b) = x^{2} + (a + b) x + ab

What each letter means
$x$	the shared term
$a$	the number added in the first bracket (negative for a minus)
$b$	the number added in the second bracket (negative for a minus)

When two brackets share the same first term x, the product is x², plus (the sum of the two numbers) times x, plus their product. Read backwards, it factorises x² + px + q: find two numbers that add to p and multiply to q.

Why it works

(x + a)(x + b) = x² + bx + ax + ab, and the two middle terms combine to (a + b)x. With algebra tiles, an x-by-x square, a strips of x, b strips of x and a-by-b unit squares fill an (x + a)-by-(x + b) rectangle.

When to use it

Multiplying brackets like (x + 3)(x − 7) quickly, factorising a quadratic whose x² term is 1, or multiplying numbers such as 103 × 104.

How to use it

To multiply, add the two numbers for the x term and multiply them for the last term.
To factorise x² + px + q, find two numbers that add to p and multiply to q.
Write (x + first)(x + second), and check by multiplying out.

To remember: Add for the middle, multiply for the end.

Other forms

Special case $(x + a)^{2} = x^{2} + 2 a x + a^{2}$ When a = b: the square of a sum.
Special case $(x + a) (x - a) = x^{2} - a^{2}$ When b = −a: the difference of squares.

Worked examples

Expand

Expand (x + 3)(x + 4).

$(x + 3) (x + 4) = x^{2} + (3 + 4) x + 3 \times 4$
$= x^{2} + 7 x + 12$

Answer: x² + 7x + 12

Factorise

Factorise x² − 5x + 6.

We need two numbers that add to −5 and multiply to 6: they are −2 and −3.
$x^{2} - 5 x + 6 = (x - 2) (x - 3)$

Answer: (x − 2)(x − 3)

Simplify

Simplify (y + 5)(y − 2) − (y + 1)(y − 1).

$(y + 5) (y - 2) = y^{2} + 3 y - 10$
$(y + 1) (y - 1) = y^{2} - 1$
$(y^{2} + 3 y - 10) - (y^{2} - 1) = 3 y - 9$

Answer: 3y − 9

Prove

Simplify (x² − 7x + 12) ÷ (5x² + 5x − 100), for values where the bottom is not 0.

Try it first, then show the working

x² − 7x + 12: two numbers adding to −7 and multiplying to 12 are −3 and −4, so it is (x − 3)(x − 4).
5x² + 5x − 100 = 5(x² + x − 20), and two numbers adding to 1 and multiplying to −20 are 5 and −4, so it is 5(x − 4)(x + 5).
Cancel the common factor (x − 4): the answer is (x − 3) ÷ 5(x + 5).

Answer: (x − 3)/(5(x + 5))

Quick calculation

Find 103 × 104 without a calculator.

Try it first, then show the working

$103 \times 104 = (100 + 3) (100 + 4)$
$= 10 0^{2} + (3 + 4) \times 100 + 3 \times 4$
$= 10000 + 700 + 12 = 10712$

Answer: 10712

Common mistake: Getting the signs wrong: in x² − 5x + 6 both numbers are negative (−2 and −3), because they add to −5 but multiply to +6.

Class 9

Product of (ax + b) and (cx + d)

(a x + b) (c x + d) = a c x^{2} + (a d + b c) x + b d

What each letter means
$x$	the variable
$a$	the number multiplying x in the first bracket
$b$	the number added in the first bracket
$c$	the number multiplying x in the second bracket
$d$	the number added in the second bracket

Multiplying two brackets of the form (ax + b) and (cx + d) gives an x² term, an x term and a number: the firsts multiply, the outer and inner products add, and the lasts multiply.

Why it works

Multiply each term of the first bracket by each term of the second: ax × cx = acx², ax × d = adx, b × cx = bcx, b × d = bd. The two middle products are both x terms, so they combine to (ad + bc)x.

When to use it

Multiplying any two linear brackets, and working backwards to factorise a quadratic whose x² term is not 1 (like 6x² + 11x + 3).

How to use it

Multiply the first terms: acx².
Multiply outer and inner, and add: (ad + bc)x.
Multiply the last terms: bd.

To remember: Firsts, outers plus inners, lasts.

Other forms

Special case $(x + b) (x + d) = x^{2} + (b + d) x + b d$ When a = c = 1: the product of (x + a) and (x + b).

Worked examples

Expand

Expand (2x + 3)(3x + 4).

$(2 x + 3) (3 x + 4) = 6 x^{2} + (2 \times 4 + 3 \times 3) x + 12$
$= 6 x^{2} + 17 x + 12$

Answer: 6x² + 17x + 12

Factorise

Factorise 6x² + 11x + 3.

We need ac = 6, bd = 3 and ad + bc = 11. Try a = 2, c = 3, b = 3, d = 1: then ad + bc = 2 + 9 = 11.
$6 x^{2} + 11 x + 3 = (2 x + 3) (3 x + 1)$

Answer: (2x + 3)(3x + 1)

Simplify

Simplify (3y − 2)(2y + 5) − 6y².

$(3 y - 2) (2 y + 5) = 6 y^{2} + 15 y - 4 y - 10 = 6 y^{2} + 11 y - 10$
$6 y^{2} + 11 y - 10 - 6 y^{2} = 11 y - 10$

Answer: 11y − 10

Prove

A rectangular pool has an area of 2x² + 7x + 3 square hastas and a width of 2x + 1 hastas. Find its length.

Try it first, then show the working

Factorise the area: we need two brackets whose product is 2x² + 7x + 3 and one of which is (2x + 1).
$(2 x + 1) (x + 3) = 2 x^{2} + 6 x + x + 3 = 2 x^{2} + 7 x + 3$
So the length is x + 3 hastas.

Answer: x + 3 hastas

Quick calculation

Find 21 × 32 by writing it as (2x + 1)(3x + 2) with x = 10.

Try it first, then show the working

$(2 x + 1) (3 x + 2) = 6 x^{2} + 7 x + 2$
$6 \times 100 + 7 \times 10 + 2 = 672$

Answer: 672

Common mistake: Forgetting one of the two middle products, or adding a and c instead of multiplying them for the x² term.

Class 9

Cube of a sum

(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}

What each letter means
$a$	the first term
$b$	the second term

Cubing a sum gives four parts: a³, three lots of a²b, three lots of ab², and b³.

Why it works

(a + b)³ = (a + b)(a + b)² = (a + b)(a² + 2ab + b²). Multiplying out gives a³ + 2a²b + ab² + a²b + 2ab² + b³, which collects to a³ + 3a²b + 3ab² + b³. As a picture, a cube of side a + b splits into a cube a³, a cube b³, three a-by-a-by-b blocks and three a-by-b-by-b blocks.

When to use it

Expanding a cubed bracket, recognising an expression as a perfect cube (the side of a cube from its volume), or cubing numbers like 101.

How to use it

Cube the first term.
Add 3 × (first)² × second.
Add 3 × first × (second)².
Add the cube of the second term.

To remember: The numbers 1, 3, 3, 1, with the power of a going down and the power of b going up.

Other forms

Same formula, another form $(a + b)^{3} = a^{3} + b^{3} + 3 ab (a + b)$ The shorter form, handy when you know a + b and ab.
Rearranged $a^{3} + b^{3} = (a + b)^{3} - 3 ab (a + b)$ The sum of two cubes from a sum and a product.

Worked examples

Expand

Expand (x + 2)³.

$(x + 2)^{3} = x^{3} + 3 x^{2} (2) + 3 x (2)^{2} + 2^{3}$
$= x^{3} + 6 x^{2} + 12 x + 8$

Answer: x³ + 6x² + 12x + 8

Factorise

The volume of a cube is p³ + 6p²q + 12pq² + 8q³ cubic units. What is its side?

Compare with a³ + 3a²b + 3ab² + b³: with a = p and b = 2q, 3a²b = 6p²q, 3ab² = 12pq² and b³ = 8q³.
$p^{3} + 6 p^{2} q + 12 p q^{2} + 8 q^{3} = (p + 2 q)^{3}$
So the side is p + 2q units.

Answer: p + 2q units

Simplify

Simplify (a + b)³ − (a − b)³.

$(a + b)^{3} - (a - b)^{3} = (a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}) - (a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3})$
$= 6 a^{2} b + 2 b^{3}$

Answer: 6a²b + 2b³

Prove

If a + b = 5 and ab = 6, find a³ + b³.

Try it first, then show the working

a³ + b³ = (a + b)³ − 3ab(a + b)
$5^{3} - 3 \times 6 \times 5 = 125 - 90 = 35$

Answer: 35

Quick calculation

Find 101³ without a calculator.

Try it first, then show the working

$10 1^{3} = (100 + 1)^{3}$
$= 10 0^{3} + 3 (100)^{2} (1) + 3 (100) (1)^{2} + 1^{3}$
$= 1000000 + 30000 + 300 + 1 = 1030301$

Answer: 1030301

Common mistake: Writing (a + b)³ = a³ + b³. Test: (1 + 1)³ = 8, but 1 + 1 = 2.

Class 9

Cube of a difference

(a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}

What each letter means
$a$	the first term
$b$	the term being taken away

Cubing a difference gives the same four parts as cubing a sum, with the signs taking turns: plus, minus, plus, minus.

Why it works

Put −b in place of b in (a + b)³ = a³ + 3a²b + 3ab² + b³. Odd powers of −b are negative and even powers are positive, so the second and fourth terms turn negative.

When to use it

Expanding a cubed bracket with a minus, recognising a perfect cube with alternating signs, or cubing numbers just below a round one (99 = 100 − 1).

How to use it

Cube the first term.
Subtract 3 × (first)² × second.
Add 3 × first × (second)².
Subtract the cube of the second term.

To remember: 1, 3, 3, 1 with alternating signs.

Other forms

Same formula, another form $(a - b)^{3} = a^{3} - b^{3} - 3 ab (a - b)$ The shorter form.

Worked examples

Expand

Expand (2x − 3)³.

$(2 x - 3)^{3} = (2 x)^{3} - 3 (2 x)^{2} (3) + 3 (2 x) (3)^{2} - 3^{3}$
$= 8 x^{3} - 36 x^{2} + 54 x - 27$

Answer: 8x³ − 36x² + 54x − 27

Factorise

Factorise 8n³ − 60n²m + 150nm² − 125m³.

8n³ is (2n)³ and 125m³ is (5m)³; then 3(2n)²(5m) = 60n²m and 3(2n)(5m)² = 150nm², with alternating signs.
$8 n^{3} - 60 n^{2} m + 150 n m^{2} - 125 m^{3} = (2 n - 5 m)^{3}$

Answer: (2n − 5m)³

Simplify

Simplify (x − 1)³ + 3(x − 1)² + 3(x − 1) + 1.

With a = x − 1 and b = 1 this is a³ + 3a²b + 3ab² + b³ = (a + b)³.
$(x - 1 + 1)^{3} = x^{3}$

Answer: x³

Prove

Show that (a − b)³ = a³ − b³ − 3ab(a − b).

Try it first, then show the working

$a^{3} - b^{3} - 3 ab (a - b) = a^{3} - b^{3} - 3 a^{2} b + 3 a b^{2}$
$= a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3} = (a - b)^{3}$

Answer: Both sides equal a³ − 3a²b + 3ab² − b³

Quick calculation

Find 99³ without a calculator.

Try it first, then show the working

$9 9^{3} = (100 - 1)^{3}$
$= 10 0^{3} - 3 (100)^{2} (1) + 3 (100) (1)^{2} - 1^{3}$
$= 1000000 - 30000 + 300 - 1 = 970299$

Answer: 970299

Common mistake: Making every term after the first negative. The signs alternate: +, −, +, −.

Class 9

Difference of two cubes

a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})

What each letter means
$a$	the term whose cube comes first
$b$	the term whose cube is taken away

A difference of two cubes always has a − b as a factor; the other factor is a² + ab + b².

Why it works

Multiply out (a − b)(a² + ab + b²) = a³ + a²b + ab² − a²b − ab² − b³. Everything in the middle cancels, leaving a³ − b³.

When to use it

Factorising an expression of the form (cube) − (cube), simplifying a fraction with a³ − b³ in it, or working out differences of cubes of numbers.

How to use it

Check both terms are cubes with a minus between them.
Find what is cubed in each (a and b).
Write (a − b)(a² + ab + b²).

To remember: Same sign in the first bracket, opposite sign for ab, and b² always plus.

Other forms

Rearranged $a^{3} - b^{3} = (a - b)^{3} + 3 ab (a - b)$ The difference of cubes from a difference and a product.

Worked examples

Expand

Multiply (x − 2)(x² + 2x + 4).

$(x - 2) (x^{2} + 2 x + 4) = x^{3} + 2 x^{2} + 4 x - 2 x^{2} - 4 x - 8$
$= x^{3} - 8$

Answer: x³ − 8

Factorise

Factorise 27y³ − 8.

27y³ is (3y)³ and 8 is 2³.
$27 y^{3} - 8 = (3 y - 2) (9 y^{2} + 6 y + 4)$

Answer: (3y − 2)(9y² + 6y + 4)

Simplify

Simplify (a³ − b³) ÷ (a − b), where a is not equal to b.

a³ − b³ = (a − b)(a² + ab + b²), and the common factor a − b cancels.

Answer: a² + ab + b²

Prove

Show that a³ − b³ = (a − b)³ + 3ab(a − b).

Try it first, then show the working

$(a - b)^{3} + 3 ab (a - b) = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3} + 3 a^{2} b - 3 a b^{2}$
$= a^{3} - b^{3}$

Answer: Both sides equal a³ − b³

Quick calculation

Find 11³ − 10³ without cubing either number.

Try it first, then show the working

$1 1^{3} - 1 0^{3} = (11 - 10) (1 1^{2} + 11 \times 10 + 1 0^{2})$
$= 1 \times (121 + 110 + 100) = 331$

Answer: 331

Common mistake: Writing the second factor as (a + b)² or as a² − ab + b². For a difference of cubes the middle term is +ab.

Class 9

Sum of two cubes

a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})

What each letter means
$a$	the first term being cubed
$b$	the second term being cubed

A sum of two cubes always has a + b as a factor; the other factor is a² − ab + b².

Why it works

Multiply out (a + b)(a² − ab + b²) = a³ − a²b + ab² + a²b − ab² + b³. The middle terms cancel, leaving a³ + b³. It is also the difference of cubes with −b in place of b.

When to use it

Factorising an expression of the form (cube) + (cube), or simplifying a fraction with a³ + b³ in it.

How to use it

Check both terms are cubes with a plus between them.
Find what is cubed in each (a and b).
Write (a + b)(a² − ab + b²).

To remember: Same sign in the first bracket, opposite sign for ab, and b² always plus.

Other forms

Rearranged $a^{3} + b^{3} = (a + b)^{3} - 3 ab (a + b)$ The sum of cubes from a sum and a product.

Worked examples

Expand

Multiply (x + 3)(x² − 3x + 9).

$(x + 3) (x^{2} - 3 x + 9) = x^{3} - 3 x^{2} + 9 x + 3 x^{2} - 9 x + 27$
$= x^{3} + 27$

Answer: x³ + 27

Factorise

Factorise 8m³ + 125.

8m³ is (2m)³ and 125 is 5³.
$8 m^{3} + 125 = (2 m + 5) (4 m^{2} - 10 m + 25)$

Answer: (2m + 5)(4m² − 10m + 25)

Simplify

Simplify (a³ + b³) ÷ (a + b), where a + b is not 0.

a³ + b³ = (a + b)(a² − ab + b²), and the common factor a + b cancels.

Answer: a² − ab + b²

Prove

Show that a³ + b³ = (a + b)³ − 3ab(a + b).

Try it first, then show the working

$(a + b)^{3} - 3 ab (a + b) = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3} - 3 a^{2} b - 3 a b^{2}$
$= a^{3} + b^{3}$

Answer: Both sides equal a³ + b³

Quick calculation

Find 10³ + 2³ as a product.

Try it first, then show the working

$1 0^{3} + 2^{3} = (10 + 2) (1 0^{2} - 10 \times 2 + 2^{2})$
$= 12 \times 84 = 1008$

Answer: 1008

Common mistake: Using +ab in the second factor. For a sum of cubes the middle term is −ab.

Class 9

a³ + b³ + c³ − 3abc

a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a)

What each letter means
$a$	the first term
$b$	the second term
$c$	the third term

The sum of three cubes minus three times their product always has a + b + c as a factor. So when a + b + c = 0, the sum of the cubes equals 3abc.

Why it works

Multiply (a + b + c) by (a² + b² + c² − ab − bc − ca) term by term: every mixed term such as a²b appears once with a plus and once with a minus and cancels, the three abc terms add to −3abc, and a³, b³ and c³ are left.

When to use it

Factorising a³ + b³ + c³ − 3abc, finding a sum of cubes from a + b + c, ab + bc + ca and abc, or using the shortcut when a + b + c = 0.

How to use it

Find a + b + c.
Find a² + b² + c² − ab − bc − ca (use (a + b + c)² to get the pair products if needed).
Multiply the two, then add 3abc if you want a³ + b³ + c³ on its own.

To remember: If they add to zero, the cubes add to three times the product.

Other forms

Special caseIf a + b + c = 0, then a³ + b³ + c³ = 3abc.When a + b + c = 0.
Same formula, another form $a^{3} + b^{3} + c^{3} - 3 ab c = \frac{1}{2} (a + b + c) ((a - b)^{2} + (b - c)^{2} + (c - a)^{2})$ The second factor is half the sum of three squares, so it is never negative.

Worked examples

Exam

Three numbers add to 10, multiply to 25, and the sum of their squares is 38. Find the sum of their cubes.

From (x + y + z)² = x² + y² + z² + 2(xy + yz + zx): 100 = 38 + 2(xy + yz + zx), so xy + yz + zx = 31.
x³ + y³ + z³ = 3xyz + (x + y + z)(x² + y² + z² − (xy + yz + zx))
$3 \times 25 + 10 \times (38 - 31) = 75 + 70 = 145$

Answer: 145

Expand

Multiply (x + y + 1)(x² + y² + 1 − xy − y − x).

This is (a + b + c)(a² + b² + c² − ab − bc − ca) with a = x, b = y, c = 1.
The product is x³ + y³ + 1 − 3xy.

Answer: x³ + y³ + 1 − 3xy

Factorise

Factorise 8x³ + y³ + 27z³ − 18xyz.

8x³ = (2x)³, y³ = y³, 27z³ = (3z)³, and 3(2x)(y)(3z) = 18xyz.
So it is (2x + y + 3z)((2x)² + y² + (3z)² − 2xy − 3yz − 6zx).

Answer: (2x + y + 3z)(4x² + y² + 9z² − 2xy − 3yz − 6zx)

Simplify

Simplify (x − y)³ + (y − z)³ + (z − x)³.

Try it first, then show the working

The three terms add to (x − y) + (y − z) + (z − x) = 0.
So their cubes add to 3 times their product: 3(x − y)(y − z)(z − x).

Answer: 3(x − y)(y − z)(z − x)

Prove

If a + b + c = 5 and ab + bc + ca = 10, prove that a³ + b³ + c³ − 3abc = −25.

Try it first, then show the working

a² + b² + c² = (a + b + c)² − 2(ab + bc + ca) = 25 − 20 = 5.
So a² + b² + c² − ab − bc − ca = 5 − 10 = −5.
$5 \times (- 5) = - 25$

Answer: −25

Quick calculation

Find 3³ + 4³ + 5³ − 3 × 3 × 4 × 5 using the identity.

Try it first, then show the working

$(3 + 4 + 5) (3^{2} + 4^{2} + 5^{2} - 3 \times 4 - 4 \times 5 - 5 \times 3) = 12 \times (50 - 47)$
$= 12 \times 3 = 36$

Answer: 36

Common mistake: Forgetting the −3abc: the factor rule is for a³ + b³ + c³ − 3abc, not for a³ + b³ + c³ alone.