If $ x^{2}+\frac{1}{x^{2}}=98 $, find the value of $ x^{3}+\frac{1}{x^{3}} $.

Given:

\( x^{2}+\frac{1}{x^{2}}=98 \)

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,

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

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

$=98+2$

$=100$

$=(10)^{2}$

$\Rightarrow x+\frac{1}{x}=10$

Cubing both sides, we get,

$(x+\frac{1}{x})^{3}=(10)^{3}$

$\Rightarrow x^{3}+\frac{1}{x^{3}}+3 \times x \times \frac{1}{x}(x+\frac{1}{x})=1000$

$\Rightarrow x^{3}+\frac{1}{x^{3}}+3 \times 10=1000$

$\Rightarrow x^{3}+\frac{1}{x^{3}}+30=1000$

$\Rightarrow x^{3}+\frac{1}{x^{3}}=1000-30$

$\Rightarrow x^{3}+\frac{1}{x^{3}}=970$

The value of \( x^{3}+\frac{1}{x^{3}} \) is $970$. 

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Simply Easy Learning

Updated on: 10-Oct-2022

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