Find the value of expression. $\frac{4}{\sqrt[3]{9}-\sqrt[3]{3}+1}$ is-

Explanation

Given that, expression $\frac{4}{\sqrt[3]{9}-\sqrt[3]{3}+1}$ $= \frac{3+1}{(3^2)^{\frac{1}{3}}-(3)^{\frac{1}{3}}+1}$ $= \frac{(3^{\frac{1}{3}})^3+(1^{\frac{1}{3}})^3}{(3^{\frac{1}{3}})^2-(3)^{\frac{1}{3}} \times 1^{\frac{1}{3}}+(1^{\frac{1}{3}})^2}$ $= \frac{(3^{\frac{1}{3}}+1^{\frac{1}{3}})[(3^{\frac{1}{3}})^2-(3)^{\frac{1}{3}} \times 1^{\frac{1}{3}}+(1^{\frac{1}{3}})^2]}{(3^{\frac{1}{3}})^2-(3)^{\frac{1}{3}} \times 1^{\frac{1}{3}}+(1^{\frac{1}{3}})^2}$ [$\because a^3+b^3=(a+b)(a^2-ab+b^2)$] $= 3^{\frac{1}{3}}+1$