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If $ x^{4}+\frac{1}{x^{4}}=119 $, find the value of $ x^{3}-\frac{1}{x^{3}} $.
Given:
\( x^{4}+\frac{1}{x^{4}}=119 \)
To do:
We have to find the value of \( x^{3}-\frac{1}{x^{3}} \).
Solution:
$x^{4}+\frac{1}{x^{4}}=119$
Adding 2 to both sides, we get,
$x^{4}+\frac{1}{x^{4}}+2=119+2$
$(x^{2})^{2}+(\frac{1}{x^{2}})^{2}+2 \times x^2 \times \frac{1}{x^2}=121$
$(x^{2}+\frac{1}{x^{2}})^{2}=(11)^{2}$
$\Rightarrow x^{2}+\frac{1}{x^{2}}=11$
Similarly,
$x^{2}+\frac{1}{x^{2}}=11$
Subtracting 2 from both sides, we get,
$x^{2}+\frac{1}{x^{2}}-2=11-2$
$x^{2}+\frac{1}{x^{2}}-2\times x \times \frac{1}{x}=9$
$(x-\frac{1}{x})^{2}=(3)^{2}$
$\Rightarrow x-\frac{1}{x}=3$
Similarly,
$x-\frac{1}{x}=3$
Cubing both sides, we get,
$(x-\frac{1}{x})^{3}=(3)^{3}$
$x^{3}-\frac{1}{x^{3}}-3 \times x \times \frac{1}{x}(x-\frac{1}{x})=27$
$x^{3}-\frac{1}{x^{3}}-3 \times 3=27$
$x^{3}-\frac{1}{x^{3}}-9=27$
$x^{3}-\frac{1}{x^{3}}=27+9$
$x^{3}-\frac{1}{x^{3}}=36$
Hence, the value of \( x^{3}-\frac{1}{x^{3}} \) is 36.