Sum and difference of cubes

The last group of iden­ti­ties turns things round. Instead of expand­ing a cube, we start with a sum or dif­fer­ence of cubes, such as $a^3 + b^3$, and write it in other forms. These results are used to fac­torise, to find val­ues quickly, and to eval­u­ate sums of cubes with­out cub­ing any­thing. The les­son ends with a remark­able fact: if three num­bers add up to zero, the sum of their cubes is three times their prod­uct.

Iden­tity 13: the sum of two cubes#

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

Proof. The first form comes straight from iden­tity 11, $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$, by tak­ing $3ab(a + b)$ to the other side.

For the sec­ond form, take out the com­mon fac­tor $(a + b)$:

$$\begin{aligned} a^3 + b^3 &= (a + b)[(a + b)^2 - 3ab] \\ &= (a + b)(a^2 + 2ab + b^2 - 3ab) \\ &= (a + b)(a^2 - ab + b^2) \end{aligned}$$

Iden­tity 14: the dif­fer­ence of two cubes#

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

Proof. From iden­tity 12, $(a - b)^3 = a^3 - b^3 - 3ab(a - b)$, so $a^3 - b^3 = (a - b)^3 + 3ab(a - b)$. Tak­ing out $(a - b)$:

$$\begin{aligned} a^3 - b^3 &= (a - b)[(a - b)^2 + 3ab] \\ &= (a - b)(a^2 - 2ab + b^2 + 3ab) \\ &= (a - b)(a^2 + ab + b^2) \end{aligned}$$

A way to remem­ber the signs: in $a^3 + b^3$ the bracket $(a + b)$ has the same sign, and the $ab$ in the sec­ond bracket has the oppo­site sign. The same holds for $a^3 - b^3$.

Exam­ple 1#

Find $(x + 2)(x^2 - 2x + 4)$.

This is $(a + b)(a^2 - ab + b^2)$ with $a = x$ and $b = 2$, since $ab = 2x$ and $b^2 = 4$.

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

Exam­ple 2#

Find $(3a - 2b)(9a^2 + 6ab + 4b^2)$.

This is $(a - b)(a^2 + ab + b^2)$ with $3a$ and $2b$ in place of $a$ and $b$: $(3a)^2 = 9a^2$, $(3a)(2b) = 6ab$ and $(2b)^2 = 4b^2$.

$$(3a - 2b)(9a^2 + 6ab + 4b^2) = (3a)^3 - (2b)^3 = 27a^3 - 8b^3$$

Exam­ple 3#

If $a + b = 10$ and $ab = 21$, find $a^3 + b^3$.

First way.

$$a^3 + b^3 = (a + b)^3 - 3ab(a + b) = 1000 - 3 \times 21 \times 10 = 1000 - 630 = 370$$

Sec­ond way. First $a^2 + b^2 = (a + b)^2 - 2ab = 100 - 42 = 58$, so $a^2 - ab + b^2 = 58 - 21 = 37$. Then

$$a^3 + b^3 = (a + b)(a^2 - ab + b^2) = 10 \times 37 = 370$$

Exam­ple 4#

Find the value of $64x^3 - 125z^3$ if $4x - 5z = 16$ and $xz = 12$.

Notice that $64x^3 = (4x)^3$ and $125z^3 = (5z)^3$. Use iden­tity 14 with $a = 4x$ and $b = 5z$, so $ab = 20xz = 240$.

$$\begin{aligned} 64x^3 - 125z^3 &= (4x - 5z)^3 + 3(4x)(5z)(4x - 5z) \\ &= 16^3 + 3 \times 240 \times 16 \\ &= 4096 + 11520 \\ &= 15616 \end{aligned}$$

Iden­tity 15: three cubes#

$$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$$

Proof. Mul­ti­ply out the right-hand side, one term of the first bracket at a time.

$$\begin{aligned} a \times (\ldots) &= a^3 + ab^2 + ac^2 - a^2b - abc - a^2c \\ b \times (\ldots) &= a^2b + b^3 + bc^2 - ab^2 - b^2c - abc \\ c \times (\ldots) &= a^2c + b^2c + c^3 - abc - bc^2 - ac^2 \end{aligned}$$

Now add the three lines. Every term like $ab^2$ appears once with a plus sign and once with a minus sign, so it can­cels. What is left is $a^3 + b^3 + c^3 - 3abc$, which is the left-hand side.

Iden­tity 16: when three num­bers add up to zero#

If $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$.

Proof. If $a + b + c = 0$, the right-hand side of iden­tity 15 is $0 \times (\ldots) = 0$. So $a^3 + b^3 + c^3 - 3abc = 0$, which gives $a^3 + b^3 + c^3 = 3abc$.

Exam­ple 5#

Find $(-12)^3 + 7^3 + 5^3$ with­out cub­ing.

The three num­bers add up to $-12 + 7 + 5 = 0$. So

$$(-12)^3 + 7^3 + 5^3 = 3 \times (-12) \times 7 \times 5 = -1260$$

Check: $-1728 + 343 + 125 = -1260$.

Exam­ple 6#

Sim­plify $(x - y)^3 + (y - z)^3 + (z - x)^3$.

The three brack­ets add up to $(x - y) + (y - z) + (z - x) = 0$. So by iden­tity 16,

$$(x - y)^3 + (y - z)^3 + (z - x)^3 = 3(x - y)(y - z)(z - x)$$

Exam­ple 7#

If $a + b + c = 6$, $ab + bc + ca = 11$ and $a^2 + b^2 + c^2 = 14$, find $a^3 + b^3 + c^3 - 3abc$.

$$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = 6 \times (14 - 11) = 18$$

Check with $a = 1$, $b = 2$, $c = 3$: $1 + 8 + 27 - 3 \times 6 = 36 - 18 = 18$.

All the iden­ti­ties together#

Iden­tity 10, the tri­no­mial square with neg­a­tive terms, fol­lows from iden­tity 9 by chang­ing signs.

Prac­tice#

1. Find $(x + 3)(x^2 - 3x + 9)$.
2. Find $(2a - 1)(4a^2 + 2a + 1)$.
3. Find $(p + 2q)(p^2 - 2pq + 4q^2)$.
4. If $a + b = 6$ and $ab = 8$, find $a^3 + b^3$.
5. If $a - b = 3$ and $ab = 10$, find $a^3 - b^3$.
6. Find $27x^3 - 8y^3$ if $3x - 2y = 5$ and $xy = 2$.
7. Find $15^3 - 8^3 - 7^3$ with­out cub­ing.
8. If $a + b + c = 0$ and $abc = 4$, find $a^3 + b^3 + c^3$.
9. If $a + b + c = 9$ and $ab + bc + ca = 26$, find $a^2 + b^2 + c^2$ and then $a^3 + b^3 + c^3 - 3abc$.

Answers#