# 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.

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Updated on: 10-Oct-2022

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