How do I prove $a^3-b^3 \geq a^2b - b^2a$ Given that $a>b>0$, prove that $a^3-b^3 \geq a^2b - b^2a$.
I have considered difference of cubes, where $a^3-b^3 = (a-b)(a^2+ab+b^2)$. However this doesn't seem to get me that far - especially when working backwards from the statement I need to prove - where I factorised the right hand side into $ab(a-b)$ and equated with the expansion earlier, giving the following inequality: $a^2+ab+b^2 \geq ab$. However by AM-GM, it follows that $a^2+b^2 \geq 2ab$ which my statement above does not follow. Help would be much appreciated.

Alternative approach:
$0 < (a - b)^3 = (a^3 - b^3) - (3)(a^2b - ab^2) \implies$
$(a^3 - b^3) > 3(a^2b - ab^2).$
Further, since $a > b,$
you know that $a^2b > ab^2 \implies (a^2b - ab^2) > 0.$
Therefore $3(a^2b - ab^2) > (a^2b - ab^2)$.
