If $ \sqrt[3]{3\left(\sqrt[3]{x}-\frac{1}{\sqrt[3]{x}}\right)}=2 $, then the value of $ \left(x-\frac{1}{x}\right) $ isA. $ \frac{728}{9} $
B. $ \frac{520}{27} $
C. $ \frac{728}{27} $
D. $ \frac{328}{15} $

Given:

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

To do:

We have to find the value of \( \left(x-\frac{1}{x}\right) \).

Solution:

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

Cubing on both sides, we get,

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

$3(\sqrt[3]{x}-\frac{1}{\sqrt[3]{x}})=8$

$(\sqrt[3]{x}-\frac{1}{\sqrt[3]{x}})=\frac{8}{3}$
Let $\sqrt[3]{x}=k$
This implies,

$k-\frac{1}{k}=\frac{8}{3}$

$\frac{k(k)-1}{k}=\frac{8}{3}$

$3(k^2-1)=8(k)$

$3k^2-8k-3=0$

$3k^2-9k+k-3=0$

$3k(k-3)+1(k-3)=0$

$(3k+1)(k-3)=0$

$3k+1=0$ or $k-3=0$

$3k=-1$ or $k=3$

$k=\frac{-1}{3}$ or $k=3$

Therefore,

$\sqrt[3]{x}=\frac{-1}{3}$ or $\sqrt[3]{x}=3$

Cubing on both sides, we get,

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

$x=\frac{-1}{27}$ or $x=27$

Now,

When $x=\frac{-1}{27}$

$x-\frac{1}{x}=\frac{-1}{27}-\frac{1}{\frac{-1}{27}}$

$=27-\frac{1}{27}$

$=\frac{27(27)-1}{27}$

$=\frac{729-1}{27}$

$=\frac{728}{27}$

When $x=27$

$x-\frac{1}{x}=27-\frac{1}{27}$

$=\frac{27(27)-1}{27}$

$=\frac{729-1}{27}$

$=\frac{728}{27}$

Option C is the correct answer.

Tutorialspoint

Updated on: 10-Oct-2022

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