The square identities have bigger relatives. One squares a sum of three terms; the others deal with cubes. Each can be read backwards to factorise an expression, just as you did with the square identities. This lesson covers the square of three terms and the four cube identities, and shows how to recognise each shape.
The square of three terms#
Expanding ( a + b + c ) 2 (a + b + c)^{2} ( a + b + c ) 2
( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 a b + 2 b c + 2 c a (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ab + 2 b c + 2 c a
So an expression with three square terms and three "twice the product" terms may be a perfect square of three terms. The signs of the product terms tell you the signs inside the bracket. For instance, if a a a − a -a − a 2 a b 2ab 2 ab 2 c a 2ca 2 c a 2 b c 2bc 2 b c
a 2 + b 2 + c 2 − 2 a b + 2 b c − 2 c a = ( − a + b + c ) 2 a^{2} + b^{2} + c^{2} - 2ab + 2bc - 2ca = (-a + b + c)^{2} a 2 + b 2 + c 2 − 2 ab + 2 b c − 2 c a = ( − a + b + c ) 2
Example 1#
Factorise x 2 + 4 y 2 + 9 z 2 + 4 x y + 12 y z + 6 z x x^{2} + 4y^{2} + 9z^{2} + 4xy + 12yz + 6zx x 2 + 4 y 2 + 9 z 2 + 4 x y + 12 y z + 6 z x
The square terms are x 2 x^{2} x 2 ( 2 y ) 2 (2y)^{2} ( 2 y ) 2 ( 3 z ) 2 (3z)^{2} ( 3 z ) 2 2 ( x ) ( 2 y ) = 4 x y 2(x)(2y) = 4xy 2 ( x ) ( 2 y ) = 4 x y 2 ( 2 y ) ( 3 z ) = 12 y z 2(2y)(3z) = 12yz 2 ( 2 y ) ( 3 z ) = 12 y z 2 ( 3 z ) ( x ) = 6 z x 2(3z)(x) = 6zx 2 ( 3 z ) ( x ) = 6 z x
x 2 + 4 y 2 + 9 z 2 + 4 x y + 12 y z + 6 z x = ( x + 2 y + 3 z ) 2 x^{2} + 4y^{2} + 9z^{2} + 4xy + 12yz + 6zx = (x + 2y + 3z)^{2} x 2 + 4 y 2 + 9 z 2 + 4 x y + 12 y z + 6 z x = ( x + 2 y + 3 z ) 2
Example 2: surds and minus signs#
Factorise 2 x 2 + y 2 + 8 z 2 − 2 2 x y + 4 2 y z − 8 z x 2x^{2} + y^{2} + 8z^{2} - 2\sqrt{2}xy + 4\sqrt{2}yz - 8zx 2 x 2 + y 2 + 8 z 2 − 2 2 x y + 4 2 y z − 8 z x
First write each square term as a square. 2 x 2 = ( 2 x ) 2 2x^{2} = (\sqrt{2}x)^{2} 2 x 2 = ( 2 x ) 2 y 2 = ( y ) 2 y^{2} = (y)^{2} y 2 = ( y ) 2 8 z 2 = ( 2 2 z ) 2 8z^{2} = (2\sqrt{2}z)^{2} 8 z 2 = ( 2 2 z ) 2 8 = 4 × 2 = 2 2 \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} 8 = 4 × 2 = 2 2
( 2 x ) 2 + ( y ) 2 + ( 2 2 z ) 2 − 2 ( 2 x ) ( y ) + 2 ( y ) ( 2 2 z ) − 2 ( 2 2 z ) ( 2 x ) (\sqrt{2}x)^{2} + (y)^{2} + (2\sqrt{2}z)^{2} - 2(\sqrt{2}x)(y) + 2(y)(2\sqrt{2}z) - 2(2\sqrt{2}z)(\sqrt{2}x) ( 2 x ) 2 + ( y ) 2 + ( 2 2 z ) 2 − 2 ( 2 x ) ( y ) + 2 ( y ) ( 2 2 z ) − 2 ( 2 2 z ) ( 2 x )
With a = 2 x a = \sqrt{2}x a = 2 x b = y b = y b = y c = 2 2 z c = 2\sqrt{2}z c = 2 2 z a 2 + b 2 + c 2 − 2 a b + 2 b c − 2 c a a^{2} + b^{2} + c^{2} - 2ab + 2bc - 2ca a 2 + b 2 + c 2 − 2 ab + 2 b c − 2 c a a a a a a a
= ( − 2 x + y + 2 2 z ) 2 = (-\sqrt{2}x + y + 2\sqrt{2}z)^{2} = ( − 2 x + y + 2 2 z ) 2
A tempting wrong answer is ( 2 x + y + 8 z ) 2 (2x + y + 8z)^{2} ( 2 x + y + 8 z ) 2
The four cube identities#
( a + b ) 3 = a 3 + b 3 + 3 a b ( a + b ) = a 3 + b 3 + 3 a 2 b + 3 a b 2 ( a − b ) 3 = a 3 − b 3 − 3 a b ( a − b ) = a 3 − b 3 − 3 a 2 b + 3 a b 2 a 3 + b 3 = ( a + b ) ( a 2 − a b + b 2 ) a 3 − b 3 = ( a − b ) ( a 2 + a b + b 2 ) \begin{aligned} (a + b)^{3} &= a^{3} + b^{3} + 3ab(a + b) = a^{3} + b^{3} + 3a^{2}b + 3ab^{2} \\ (a - b)^{3} &= a^{3} - b^{3} - 3ab(a - b) = a^{3} - b^{3} - 3a^{2}b + 3ab^{2} \\ a^{3} + b^{3} &= (a + b)(a^{2} - ab + b^{2}) \\ a^{3} - b^{3} &= (a - b)(a^{2} + ab + b^{2}) \end{aligned} ( a + b ) 3 ( a − b ) 3 a 3 + b 3 a 3 − b 3 = a 3 + b 3 + 3 ab ( a + b ) = a 3 + b 3 + 3 a 2 b + 3 a b 2 = a 3 − b 3 − 3 ab ( a − b ) = a 3 − b 3 − 3 a 2 b + 3 a b 2 = ( a + b ) ( a 2 − ab + b 2 ) = ( a − b ) ( a 2 + ab + b 2 )
The first two show what a perfect cube looks like when expanded: four terms. The last two factorise a sum and a difference of two cubes. Notice the signs in the last two: the bracket ( a + b ) (a + b) ( a + b ) − a b -ab − ab ( a − b ) (a - b) ( a − b ) + a b +ab + ab
Example 3: a perfect cube#
Factorise x 3 − 12 x ( x − 4 ) − 64 x^{3} - 12x(x - 4) - 64 x 3 − 12 x ( x − 4 ) − 64
Open the bracket first: x 3 − 12 x 2 + 48 x − 64 x^{3} - 12x^{2} + 48x - 64 x 3 − 12 x 2 + 48 x − 64 x 3 x^{3} x 3 64 = 4 3 64 = 4^{3} 64 = 4 3 ( a − b ) 3 (a - b)^{3} ( a − b ) 3 a = x a = x a = x b = 4 b = 4 b = 4
x 3 − 12 x 2 + 48 x − 64 = ( x ) 3 − ( 4 ) 3 − 3 ( x ) 2 ( 4 ) + 3 ( x ) ( 4 ) 2 = ( x − 4 ) 3 \begin{aligned} x^{3} - 12x^{2} + 48x - 64 &= (x)^{3} - (4)^{3} - 3(x)^{2}(4) + 3(x)(4)^{2} \\ &= (x - 4)^{3} \end{aligned} x 3 − 12 x 2 + 48 x − 64 = ( x ) 3 − ( 4 ) 3 − 3 ( x ) 2 ( 4 ) + 3 ( x ) ( 4 ) 2 = ( x − 4 ) 3
Both middle terms match: 3 x 2 × 4 = 12 x 2 3x^{2} \times 4 = 12x^{2} 3 x 2 × 4 = 12 x 2 3 x × 16 = 48 x 3x \times 16 = 48x 3 x × 16 = 48 x
Example 4: a sum of cubes#
Factorise 8 x 3 + 27 8x^{3} + 27 8 x 3 + 27
8 x 3 + 27 = ( 2 x ) 3 + 3 3 = ( 2 x + 3 ) [ ( 2 x ) 2 − ( 2 x ) ( 3 ) + 3 2 ] = ( 2 x + 3 ) ( 4 x 2 − 6 x + 9 ) \begin{aligned} 8x^{3} + 27 &= (2x)^{3} + 3^{3} \\ &= (2x + 3)\left[(2x)^{2} - (2x)(3) + 3^{2}\right] \\ &= (2x + 3)(4x^{2} - 6x + 9) \end{aligned} 8 x 3 + 27 = ( 2 x ) 3 + 3 3 = ( 2 x + 3 ) [ ( 2 x ) 2 − ( 2 x ) ( 3 ) + 3 2 ] = ( 2 x + 3 ) ( 4 x 2 − 6 x + 9 )
Example 5: a difference of cubes#
Factorise 27 a 3 − 64 b 3 27a^{3} - 64b^{3} 27 a 3 − 64 b 3
27 a 3 − 64 b 3 = ( 3 a ) 3 − ( 4 b ) 3 = ( 3 a − 4 b ) ( 9 a 2 + 12 a b + 16 b 2 ) \begin{aligned} 27a^{3} - 64b^{3} &= (3a)^{3} - (4b)^{3} \\ &= (3a - 4b)(9a^{2} + 12ab + 16b^{2}) \end{aligned} 27 a 3 − 64 b 3 = ( 3 a ) 3 − ( 4 b ) 3 = ( 3 a − 4 b ) ( 9 a 2 + 12 ab + 16 b 2 )
Example 6: an incomplete cube#
Factorise x 3 + 3 x 2 + 3 x − 7 x^{3} + 3x^{2} + 3x - 7 x 3 + 3 x 2 + 3 x − 7
The first three terms look like the start of ( x + 1 ) 3 = x 3 + 3 x 2 + 3 x + 1 (x + 1)^{3} = x^{3} + 3x^{2} + 3x + 1 ( x + 1 ) 3 = x 3 + 3 x 2 + 3 x + 1 − 7 -7 − 7 1 − 8 1 - 8 1 − 8
x 3 + 3 x 2 + 3 x − 7 = ( x ) 3 + 3 ( x ) 2 ( 1 ) + 3 ( x ) ( 1 ) 2 + 1 3 − 8 = ( x + 1 ) 3 − 2 3 \begin{aligned} x^{3} + 3x^{2} + 3x - 7 &= (x)^{3} + 3(x)^{2}(1) + 3(x)(1)^{2} + 1^{3} - 8 \\ &= (x + 1)^{3} - 2^{3} \end{aligned} x 3 + 3 x 2 + 3 x − 7 = ( x ) 3 + 3 ( x ) 2 ( 1 ) + 3 ( x ) ( 1 ) 2 + 1 3 − 8 = ( x + 1 ) 3 − 2 3
Now it is a difference of cubes with a = x + 1 a = x + 1 a = x + 1 b = 2 b = 2 b = 2
= ( x + 1 − 2 ) [ ( x + 1 ) 2 + ( x + 1 ) 2 + 2 2 ] = ( x − 1 ) [ x 2 + 2 x + 1 + 2 x + 2 + 4 ] = ( x − 1 ) ( x 2 + 4 x + 7 ) \begin{aligned} &= (x + 1 - 2)\left[(x + 1)^{2} + (x + 1)2 + 2^{2}\right] \\ &= (x - 1)\left[x^{2} + 2x + 1 + 2x + 2 + 4\right] \\ &= (x - 1)(x^{2} + 4x + 7) \end{aligned} = ( x + 1 − 2 ) [ ( x + 1 ) 2 + ( x + 1 ) 2 + 2 2 ] = ( x − 1 ) [ x 2 + 2 x + 1 + 2 x + 2 + 4 ] = ( x − 1 ) ( x 2 + 4 x + 7 )
Example 7: sixth powers#
Factorise x 6 + y 6 x^{6} + y^{6} x 6 + y 6
A sixth power is a cube of a square: x 6 = ( x 2 ) 3 x^{6} = (x^{2})^{3} x 6 = ( x 2 ) 3 a = x 2 a = x^{2} a = x 2 b = y 2 b = y^{2} b = y 2
x 6 + y 6 = ( x 2 ) 3 + ( y 2 ) 3 = ( x 2 + y 2 ) [ ( x 2 ) 2 − x 2 ⋅ y 2 + ( y 2 ) 2 ] = ( x 2 + y 2 ) ( x 4 − x 2 y 2 + y 4 ) \begin{aligned} x^{6} + y^{6} &= (x^{2})^{3} + (y^{2})^{3} \\ &= (x^{2} + y^{2})\left[(x^{2})^{2} - x^{2} \cdot y^{2} + (y^{2})^{2}\right] \\ &= (x^{2} + y^{2})(x^{4} - x^{2}y^{2} + y^{4}) \end{aligned} x 6 + y 6 = ( x 2 ) 3 + ( y 2 ) 3 = ( x 2 + y 2 ) [ ( x 2 ) 2 − x 2 ⋅ y 2 + ( y 2 ) 2 ] = ( x 2 + y 2 ) ( x 4 − x 2 y 2 + y 4 )
Practice#
Factorise.
4 x 2 + y 2 + z 2 + 4 x y + 2 y z + 4 z x 4x^{2} + y^{2} + z^{2} + 4xy + 2yz + 4zx 4 x 2 + y 2 + z 2 + 4 x y + 2 y z + 4 z x 9 a 2 + b 2 + 4 c 2 − 6 a b − 4 b c + 12 c a 9a^{2} + b^{2} + 4c^{2} - 6ab - 4bc + 12ca 9 a 2 + b 2 + 4 c 2 − 6 ab − 4 b c + 12 c a 8 x 3 + 36 x 2 y + 54 x y 2 + 27 y 3 8x^{3} + 36x^{2}y + 54xy^{2} + 27y^{3} 8 x 3 + 36 x 2 y + 54 x y 2 + 27 y 3 27 − 27 a + 9 a 2 − a 3 27 - 27a + 9a^{2} - a^{3} 27 − 27 a + 9 a 2 − a 3 x 3 + 125 x^{3} + 125 x 3 + 125 64 m 3 − 27 n 3 64m^{3} - 27n^{3} 64 m 3 − 27 n 3 2 x 3 + 54 2x^{3} + 54 2 x 3 + 54 x 3 − 3 x 2 + 3 x + 7 x^{3} - 3x^{2} + 3x + 7 x 3 − 3 x 2 + 3 x + 7 x 6 − y 6 x^{6} - y^{6} x 6 − y 6
Answers#
Show answers
( 2 x ) 2 + y 2 + z 2 + 2 ( 2 x ) ( y ) + 2 ( y ) ( z ) + 2 ( z ) ( 2 x ) = ( 2 x + y + z ) 2 (2x)^{2} + y^{2} + z^{2} + 2(2x)(y) + 2(y)(z) + 2(z)(2x) = (2x + y + z)^{2} ( 2 x ) 2 + y 2 + z 2 + 2 ( 2 x ) ( y ) + 2 ( y ) ( z ) + 2 ( z ) ( 2 x ) = ( 2 x + y + z ) 2 The terms with minus signs both contain b b b b b b ( 3 a − b + 2 c ) 2 (3a - b + 2c)^{2} ( 3 a − b + 2 c ) 2
( 2 x ) 3 + 3 ( 2 x ) 2 ( 3 y ) + 3 ( 2 x ) ( 3 y ) 2 + ( 3 y ) 3 = ( 2 x + 3 y ) 3 (2x)^{3} + 3(2x)^{2}(3y) + 3(2x)(3y)^{2} + (3y)^{3} = (2x + 3y)^{3} ( 2 x ) 3 + 3 ( 2 x ) 2 ( 3 y ) + 3 ( 2 x ) ( 3 y ) 2 + ( 3 y ) 3 = ( 2 x + 3 y ) 3 3 3 − 3 ( 3 ) 2 a + 3 ( 3 ) a 2 − a 3 = ( 3 − a ) 3 3^{3} - 3(3)^{2}a + 3(3)a^{2} - a^{3} = (3 - a)^{3} 3 3 − 3 ( 3 ) 2 a + 3 ( 3 ) a 2 − a 3 = ( 3 − a ) 3 ( x + 5 ) ( x 2 − 5 x + 25 ) (x + 5)(x^{2} - 5x + 25) ( x + 5 ) ( x 2 − 5 x + 25 ) ( 4 m − 3 n ) ( 16 m 2 + 12 m n + 9 n 2 ) (4m - 3n)(16m^{2} + 12mn + 9n^{2}) ( 4 m − 3 n ) ( 16 m 2 + 12 mn + 9 n 2 ) 2 ( x 3 + 27 ) = 2 ( x + 3 ) ( x 2 − 3 x + 9 ) 2(x^{3} + 27) = 2(x + 3)(x^{2} - 3x + 9) 2 ( x 3 + 27 ) = 2 ( x + 3 ) ( x 2 − 3 x + 9 ) ( x − 1 ) 3 + 8 = ( x − 1 ) 3 + 2 3 = ( x − 1 + 2 ) [ ( x − 1 ) 2 − 2 ( x − 1 ) + 4 ] = ( x + 1 ) ( x 2 − 4 x + 7 ) (x - 1)^{3} + 8 = (x - 1)^{3} + 2^{3} = (x - 1 + 2)\left[(x - 1)^{2} - 2(x - 1) + 4\right] = (x + 1)(x^{2} - 4x + 7) ( x − 1 ) 3 + 8 = ( x − 1 ) 3 + 2 3 = ( x − 1 + 2 ) [ ( x − 1 ) 2 − 2 ( x − 1 ) + 4 ] = ( x + 1 ) ( x 2 − 4 x + 7 ) ( x 3 + y 3 ) ( x 3 − y 3 ) = ( x + y ) ( x 2 − x y + y 2 ) ( x − y ) ( x 2 + x y + y 2 ) (x^{3} + y^{3})(x^{3} - y^{3}) = (x + y)(x^{2} - xy + y^{2})(x - y)(x^{2} + xy + y^{2}) ( x 3 + y 3 ) ( x 3 − y 3 ) = ( x + y ) ( x 2 − x y + y 2 ) ( x − y ) ( x 2 + x y + y 2 )