If x4 + (1/x4) = 119, Find the value of

Solution:

Given, $x^{4}+\left(1 / x^{4}\right)=119 \quad \ldots .1$

We know that $(x+y)^{2}=x^{2}+y^{2}+2 x y$

Substitute $x^{4}+\left(1 / x^{4}\right)=119$ in eq 1

$\left(x^{2}+\left(1 / x^{2}\right)\right)^{2}=x^{4}+\left(1 / x^{4}\right)+\left(2^{*} x^{2} * 1 / x^{2}\right)$

$=x^{4}+\left(1 / x^{4}\right)+2$

= 119 + 2

= 121

$\left(x^{2}+\left(1 / x^{2}\right)\right)^{2}=121$

$x^{2}+\left(\frac{1}{x^{2}}\right)=\sqrt{121}$

$x^{2}+\left(1 / x^{2}\right)=\pm 11$

Now, find $(x-1 / x)$

We know that $(x-y)^{2}=x^{2}+y^{2}-2 x y$

$(x-1 / x)^{2}=x^{2}+1 / x^{2}-\left(2^{*} x^{*} 1 / x\right.$

$=x^{2}+1 / x^{2}-2$

$=11-2$

$=9$

$(x-1 / x)=\sqrt{9}$

$=\pm 3$

We need to find $x^{3}-\left(1 / x^{3}\right)$

We know that, $a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}-a b\right)$

$x^{3}-\left(1 / x^{3}\right)=(x-1 / x)\left(x^{2}+\left(1 / x^{2}\right)+x^{*} 1 / x\right.$

Here, $x^{2}+\left(1 / x^{2}\right)=11$ and $(x-1 / x)=3$

$x^{3}-\left(1 / x^{3}\right)=3(11+1)$

= 3(12)

= 36

Hence, the value of $x^{3}-\left(1 / x^{3}\right)=36$