The Sum and Difference of Two Cubes

Squares, then cubes

A difference of two squares factors into two brackets: $a^{2} - b^{2} = \left(a - b\right)\left(a + b\right)$. A difference of two cubes, $a^{3} - b^{3}$, factors too, but the second bracket has three terms: $a^{3} - b^{3} = \left(a - b\right)\left(a^{2} + ab + b^{2}\right)$.

The first bracket, a − b, is linear: its terms have no powers. The second bracket, $a^{2} + ab + b^{2}$, is quadratic, because its highest power is 2.

Why the middle terms cancel

Check the identity by multiplying out the right side. Multiply each term of the first bracket by every term of the second.

Multiplying the second bracket by a gives $a^{3} + a^{2}b + ab^{2}$. Multiplying it by −b gives $-a^{2}b - ab^{2} - b^{3}$. Add the two rows: $a^{3} + a^{2}b + ab^{2} - a^{2}b - ab^{2} - b^{3}$.

The $a^{2}b$ terms cancel, and so do the $ab^{2}$ terms, which leaves $a^{3} - b^{3}$. The second bracket is chosen so that exactly this happens: every term that is not a cube appears once with a plus sign and once with a minus sign.

Numbers confirm it. With a = 10 and b = 2, $a^{3} - b^{3} = 1000 - 8 = 992$, and $\left(a - b\right)\left(a^{2} + ab + b^{2}\right) = 8 \times \left(100 + 20 + 4\right) = 8 \times 124 = 992$.

The sum of two cubes

For a sum of two cubes, change two signs: $a^{3} + b^{3} = \left(a + b\right)\left(a^{2} - ab + b^{2}\right)$. The linear bracket takes the sign of the sum, a plus, and the middle term of the quadratic bracket takes the opposite sign, a minus. The last term, $b^{2}$, is always a plus.

Multiplying out checks it in the same way: $a\left(a^{2} - ab + b^{2}\right) + b\left(a^{2} - ab + b^{2}\right) = a^{3} - a^{2}b + ab^{2} + a^{2}b - ab^{2} + b^{3} = a^{3} + b^{3}$.

A sum of two squares, such as $a^{2} + b^{2}$, has no factors like these. A sum of two cubes does.

Write the number as a cube first

To factor $x^{3} - 8$, first write 8 as a cube: $8 = 2^{3}$, so $x^{3} - 8 = x^{3} - 2^{3}$. Now it matches $a^{3} - b^{3}$ with a = x and b = 2. Put these into the identity: $\left(x - 2\right)\left(x^{2} + x \times 2 + 2^{2}\right) = \left(x - 2\right)\left(x^{2} + 2x + 4\right)$.

To factor $x^{3} + 27$, write $27 = 3^{3}$. Then a = x and b = 3, and the identity for a sum gives $\left(x + 3\right)\left(x^{2} - 3x + 9\right)$.

The cubes worth knowing by sight are 1, 8, 27, 64, 125, 216, 343, 512, 729 and 1000, the cubes of 1 to 10. A letter term can be a cube too: $8x^{3} = \left(2x\right)^{3}$, so $8x^{3} - 1 = \left(2x - 1\right)\left(4x^{2} + 2x + 1\right)$.

The usual mistakes

Copying the sign into the middle term. In $x^{3} + 27$ the middle term of the quadratic bracket is −3x, not +3x. The middle sign is always the opposite of the sign in the linear bracket.

Leaving out the middle term, as in $\left(x + 3\right)\left(x^{2} + 9\right)$. That treats a sum of cubes as if it were like a difference of squares. A cube always leaves three terms in the second bracket.

Writing $a^{3} - b^{3}$ as $\left(a - b\right)^{3}$. The difference of two cubes is not the cube of the difference: with a = 10 and b = 2, $\left(a - b\right)^{3} = 8^{3} = 512$, but $a^{3} - b^{3} = 992$.