Question

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

Answer 1

In the given problem, we have to find the value of $x^2+\frac{1}{x^2},x^3+\frac{1}{x^3}$ and $x^4+\frac{1}{x^4}$

Given: $x+\frac{1}{x}=3$,

We shall use the identity $(a+b{)}^2=a^2+b^2+2ab$

Here putting, $x+\frac{1}{x}=3$,

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

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

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

$\Rightarrow x^2+\frac{1}{x^2}=9-2=7$

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

Again squaring on both sides we get,

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

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

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

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

$\therefore x^4+\frac{1}{x^4}=47$...(ii)

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

Again cubing on both sides we get,

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

We shall use identity $(a+b{)}^3=a^3+b^3+3ab(a+b)$

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

$\Rightarrow x^3+\frac{1}{x^3}+3\left(3\right)=27$

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

$\therefore x^3+\frac{1}{x^3}=18$...(iii)