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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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