Question

If $\sqrt[3]{33\left(\sqrt[3]{3x}-\frac{1}{\sqrt[3]{3x}}\right)}=2,$ then $\sqrt[3]{3x}+\frac{1}{\sqrt[3]{3x}}=$

Answer 1

$\sqrt[3]{33(\sqrt[3]{3x}-\frac{1}{\sqrt[3]{3x}})}=2$

Taking cube on both the sides of the equation,

$\Rightarrow 3(x^{\frac{1}{3}}-\frac{1}{x^{\frac{1}{3}}})=8$

$\Rightarrow (x^{\frac{1}{3}}-\frac{1}{x^{\frac{1}{3}}})=\frac{8}{3}$

$\Rightarrow {(x^{\frac{1}{3}}+\frac{1}{x^{\frac{1}{3}}})}^2={(x^{\frac{1}{3}}-\frac{1}{x^{\frac{1}{3}}})}^2+4x^{\frac{1}{3}}\cdot \frac{1}{x^{\frac{1}{3}}}$

$=\frac{64}{9}+4=\frac{64+36}{9}=\frac{100}{9}$ $[\because (a+b{)}^2=(a-b{)}^2+4ab]$

$\therefore x^{\frac{1}{3}}+\frac{1}{x^{\frac{1}{3}}}=\pm \frac{10}{3}$