Question:
If $x+1 / x=3$, calculate $x^{2}+1 / x^{2}, x^{3}+1 / x^{3}, x^{4}+1 / x^{4}$
Solution:
Given, $x+1 / x=3$
We know that $(x+y)^{2}=x^{2}+y^{2}+2 x y$
$(x+1 / x)^{2}=x^{2}+1 / x^{2}+(2 * x * 1 / x)$
$3^{2}=x^{2}+1 / x^{2}+2$
$9-2=x^{2}+1 / x^{2}$
$x^{2}+1 / x^{2}=7$
Squaring on both sides
$\left(x^{2}+1 / x^{2}\right)^{2}=7^{2}$
$x^{4}+1 / x^{4}+2^{*} x^{2} * 1 / x^{2}=49$
$x^{4}+1 / x^{4}+2=49$
$x^{4}+1 / x^{4}=49-2$
$x^{4}+1 / x^{4}=47$
Again, cubing on both sides
$(x+1 / x)^{3}=3^{3}$
$x^{3}+1 / x^{3}+3 x^{*} 1 / x(x+1 / x)=27$
$x^{3}+1 / x^{3}+\left(3^{*} 3\right)=27$
$x^{3}+1 / x^{3}+9=27$
$x^{3}+1 / x^{3}=27-9$
$x^{3}+1 / x^{3}=18$
Hence, the values are $x^{2}+1 / x^{2}=7, x^{4}+1 / x^{4}=47, x^{3}+1 / x^{3}=18$