${a^3 + b^3\over a+b}$ for integer values $a$ and $b$.

Question

I am new to this website so sorry in advance if my formatting is a bit off. I am trying to show that $${a^3 + b^3\over a+b}$$ is an integer when both $a$ and $b$ are integers. I have already looked at the numbers from $1$ to $10$ for values $a$ and $b$ and it has held up for some arbitrary numbers like $99$ and $-55$, however I am unsure how to approach showing that this is true (or untrue) for all integers.

Could anybody point me in the right direction or provide a solution?

Answer 1

Recall that the sum of two cubes can be factored as $$a^3+b^3=(a+b)(a^2-ab+b^2)$$

Answer 2

HINT: use that $$a^3+b^3=(a+b)(a^2-ab+b^2)$$

Answer 3

Recall that the geometric sum is

$$\frac{1-q^n}{1-q}=1+q+\dots+q^{n-1}$$

Substitute: $q=\frac{a}{-b}$ $$\frac{1-\frac{a}{-b}^n}{1-\frac{a}{-b}}=\frac{1}{b^{n-1}}\frac{-b^n-a^n}{-b-a}=\frac{1}{b^{n-1}}\frac{a^n+b^n}{a+b}$$

Notice that $$b^{n-1}(1+q+\dots+q^{n-1})$$ is an integer if a and b are integers