Question

Simplify : $\sqrt[3]{3x^3-7x^2}>(-36{)}^{1/3}$

Answer 1

$\sqrt[3]{3x^3-7x^2}>(-36{)}^{\frac{1}{3}}$

cubing on both sides,

$x^3-7x^2>-36$

$\therefore x^3-7x^2+36>0$

$\therefore (x+2)(x-3)(x-6)>0$

$\therefore x\in (-2,3)\cup (6,\infty )$

This is the required solution.