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If \( \sqrt[3]{3\left(\sqrt[3]{x}-\frac{1}{\sqrt[3]{x}}\right)}=2 \), then the value of \( \left(x-\frac{1}{x}\right) \) is
A. \( \frac{728}{9} \)
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.