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  4. If √[3]3(√[3]x-(1/√[3]x))=2 then √[3]x+(1/√[3]x)=

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

Squares and Square Roots and Cubes and Cube Roots

Solution:

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+4 a b\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$
$ \left.\therefore\left(\sqrt[3]{x}+\frac{1}{\sqrt[3]{x}}\right)^2=\frac{1}{\sqrt[3]{x}}\right)^2=\frac{100}{9} $
$ \therefore \sqrt[3]{x}+\frac{1}{\sqrt[3]{x}}= \pm \frac{10}{3}$