Let $a, b, c$ be positive real numbers such that $a^3 + b^3 = c^3$. Prove that $a^2 + b^2 - c^2 \gt 6(c - a)(c - b)$ . 
Let $a$, $b$ and $c$ be positive real numbers such that $a^3 + b^3 = c^3$. Prove that $$a^2 + b^2 - c^2 \gt 6(c - a)(c - b).$$ 

This problem is from the Indian MO 2009, however, I do not know how to approach it, any help? 

Hint: Consider the quadratic equation $f(t) = 7t^2-6t(a+b)+6ab-a^2-b^2.$ You are to show that $f(\sqrt[3]{a^3+b^3})<0.$ But instead of plugging it in, it easier to show that the number $c = \sqrt[3]{a^3+b^3}$ lies strictly between the two roots of the quadratic $f(t).$
