If $ x^{4}+\frac{1}{x^{4}}=194 $, find $ x^{3}+\frac{1}{x^{3}}, x^{2}+\frac{1}{x^{2}} $ and $ x+\frac{1}{x} $

Given:

\( x^{4}+\frac{1}{x^{4}}=194 \)

To do:

We have to find \( x^{3}+\frac{1}{x^{3}}, x^{2}+\frac{1}{x^{2}} \) and \( x+\frac{1}{x} \).

Solution:

$x^{4}+\frac{1}{x^{4}}=194$

Adding 2 to both sides, we get,

$x^{4}+\frac{1}{x^{4}}+2=194+2$

$x^{4}+\frac{1}{x^{4}}+2=196$

$(x^{2})^{2}+\frac{1}{(x^{2})^{2}}+2\times x^2 \times \frac{1}{x^2}=(14)^{2}$

$(x^{2}+\frac{1}{x^{2}})^{2}=(14)^{2}$

$\Rightarrow x^{2}+\frac{1}{x^{2}}=14$

Similarly,

$x^{2}+\frac{1}{x^{2}}=14$
Adding 2 to both sides, we get,

$x^{2}+\frac{1}{x^{2}}+2=14+2$

$x^{2}+\frac{1}{x^{2}}+2\times x \times \frac{1}{x}=16$

$x^{2}+\frac{1}{x^{2}}+2\times x \times \frac{1}{x}=(4)^{2}$

$(x+\frac{1}{x})^{2}=(4)^{2}$

$\Rightarrow x+\frac{1}{x}=4$

Similarly,

$x+\frac{1}{x}=4$

Cubing both sides, we get,

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

$(4)^{3}=x^{3}+\frac{1}{x^{3}}+3 \times 4$

$64=x^{3}+\frac{1}{x^{3}}+12$

$x^{3}+\frac{1}{x^{3}}=64-12$

$x^{3}+\frac{1}{x^{3}}=52$

Hence, \( x^{3}+\frac{1}{x^{3}}=52, x^{2}+\frac{1}{x^{2}}=14 \) and \( x+\frac{1}{x}=4 \). 

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Updated on: 10-Oct-2022

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