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

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

40 Views