Question

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

• A. $\frac{10}{3}$
• B. $-\frac{10}{3}$
• C. $\frac{-8}{3}$
• D. $\frac{8}{3}$

Answer 1

Answer: (A), (D)

Given: $\sqrt[3]{3\left(\sqrt[3]{x}-\frac{1}{\sqrt[3]{x}}\right)}=2$
Taking cube on both sides.
${\left(\sqrt[3]{3\left(\sqrt[3]{x}-\frac{1}{\sqrt[3]{x}}\right)}\right)}^3=(2{)}^3$
$3\left(\sqrt[3]{x}-\frac{1}{\sqrt[3]{x}}\right)=8$
$\sqrt[3]{x}-\frac{1}{\sqrt[3]{x}}=\frac{8}{3}$...(1)
Now,
${\left(\sqrt[3]{x}+\frac{1}{\sqrt[3]{x}}\right)}^2={\left(\sqrt[3]{x}-\frac{1}{\sqrt[3]{x}}\right)}^2+4\cdot \sqrt[3]{x}\cdot \frac{1}{\sqrt[3]{x}}$
$\left[\text{ As, }(a+b{)}^2=(a-b{)}^2+4ab\right]$
$\therefore {\left(\sqrt[3]{x}+\frac{1}{\sqrt[3]{x}}\right)}^2={\left(\frac{8}{3}\right)}^2+4$
$\therefore {\left(\sqrt[3]{x}+\frac{1}{\sqrt[3]{x}}\right)}^2=\frac{64}{9}+4$
${\therefore {\left(\sqrt[3]{x}+\frac{1}{\sqrt[3]{x}}\right)}^2=\frac{1}{\sqrt[3]{x}})}^2=\frac{100}{9}$
$\therefore \sqrt[3]{x}+\frac{1}{\sqrt[3]{x}}=\pm \frac{10}{3}$