If $ x^{2}+\frac{1}{x^{2}}=51 $, find the value of $ x^{3}-\frac{1}{x^{3}} $.

Given:

\( x^{2}+\frac{1}{x^{2}}=51 \)

To do:

We have to find the value of \( x^{3}-\frac{1}{x^{3}} \).

Solution:

We know that,

$(a-b)^3=a^3 - b^3 - 3ab(a-b)$

Therefore,

$x^{2}+\frac{1}{x^{2}}=51$

$(x-\frac{1}{x})^{2}=x^{2}+\frac{1}{x^{2}}-2\times x \times \frac{1}{x}$

$=x^{2}+\frac{1}{x^{2}}-2$

$=51-2$

$=49$

$=(7)^{2}$

$\Rightarrow x-\frac{1}{x}=7$

Cubing both sides, we get,

$(x-\frac{1}{x})^{3}=(7)^{3}$

$\Rightarrow x^{3}-\frac{1}{x^{3}}-3(x-\frac{1}{x})=343$

$\Rightarrow x^{3}-\frac{1}{x^{3}}-3 \times 7=343$

$\Rightarrow x^{3}-\frac{1}{x^{3}}=343+21$

$\Rightarrow x^{3}-\frac{1}{x^{3}}=364$

The value of \( x^{3}-\frac{1}{x^{3}} \) is $364$.

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Updated on: 10-Oct-2022

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