Question

Find the cube root of $9a{b}^2+(b^2+24a^2)\sqrt{b^2-3a^2}$.

Answer 1

$9a{b}^2+(b^2+24a^2)\sqrt{b^2-3a^2}$

$=(b^2-3a^2)\sqrt{b^2-3a^2}+27\sqrt{b^2-3a^2}+27a^3+9a{b}^2-27a^3$

$=(3a{)}^3+(\sqrt{b^2-3a^2}{)}^3+3\times 3a(\sqrt{b^2-3a^2}{)}^2+3\times (3a{)}^2\times \sqrt{b^2-3a^2}$

$=(3a+\sqrt{b^2-3a^2}{)}^3$

Hence, cube root is $3a+\sqrt{b^2-3a^2}$