Sum of two cubes

In mathematics, the sum of two cubes is a cubed number added to another cubed number. Every sum of cubes may be factored according to the identity

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

in elementary algebra. The end term of the identity, $b^{2}$, for the sum or difference of two cubes will always end in the addition of $b^{2}$.

Proof

Starting from the left-hand side, distribute $a^{2}-ab+b^{2}$ to $a+b$ to get

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

Using the distributive law, distribute a to $a^{2}-ab+b^{2}$ and b to $a^{2}-ab+b^{2}$ to get

$a^{3}-a^{2}b+ab^{2}+ba^{2}-b^{2}a+b^{3}$

By combining, both middle terms cancel:

$-a^{2}b+ba^{2}+ab^{2}-b^{2}a=0$

leaving $a^{3}+b^{3}$

The identity does not actually equal a cube. In order to prove this, a and b must be a non-zero rational number. We will make $a=1$ and $b=2$. Plugging in a and b shows that

$a^{3}+b^{3}=(1+2)(1^{2}-1(2)+2^{2})$

Which if simplified shows that

$a^{3}+b^{3}=3(1-2+4)$

And simplifying the equation using the order of operations gets

$a^{3}+b^{3}=9$

9 is the resulting answer, although it is not a cube. However, it is a square, and so is $a=2$ and $b=2$, which equals to 16.

The only numbers that are not a sum of two cubes are 3 and 4.