Show that when n is even,

$\displaystyle

1^3-2^3+3^3-...-n^3 \equiv 1^3 + 2^3 + 3^3 +...+ n^3 - 16(1^3 + 2^3 + 3^3 +...+(\frac{n}{2})^3) \equiv \sum\limits_{r=1}^{n} r^3 - 16 \sum\limits_{r=1}^{\frac{n}{2}} r^3$

I don't know how to approach this question, the 'when n is even' is throwing me, and even without that part, I think I'd struggle. How should I begin?