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

Given:

\( x+\frac{1}{x}=3 \)

To do:

We have to calculate \( x^{2}+\frac{1}{x^{2}}, x^{3}+\frac{1}{x^{3}} \) and \( x^{4}+\frac{1}{x^{4}} \).

Solution:

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

Squaring both sides, we get,

$(x+\frac{1}{x})^{2}=(3)^{2}$

$x^{2}+\frac{1}{x^{2}}+2\times x \times \frac{1}{x}=9$

$x^{2}+\frac{1}{x^{2}}=9-2$

$x^{2}+\frac{1}{x^{2}}=7$

Similarly,

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

Cubing both sides, we get,

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

$x^{3}+\frac{1}{x^{3}}+3\times x \times \frac{1}{x} (x+\frac{1}{x})=27$

$x^{3}+\frac{1}{x^{3}}+3 \times 3=27$

$x^{3}+\frac{1}{x^{3}}+9=27$

$x^{3}+\frac{1}{x^{3}}=27-9$

$x^{3}+\frac{1}{x^{3}}=18$

Similarly,

$x^{2}+\frac{1}{x^{2}}=7$

Squaring both sides, we get,

$(x^{2}+\frac{1}{x})^{2}=(7)^{2}$

$x^{4}+\frac{1}{x^{4}}+2 \times x^2 \times \frac{1}{x^2}=49$

$x^{4}+\frac{1}{x^{4}}=49-2$

$x^{4}+\frac{1}{x^{4}}=47$

Hence, \( x^{2}+\frac{1}{x^{2}}=7, x^{3}+\frac{1}{x^{3}}=18 \) and \( x^{4}+\frac{1}{x^{4}}=47 \).

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Updated on: 10-Oct-2022

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