Question

If $x+\frac{1}{x}=2$, then find the value of $x^5+\frac{1}{x^5}$

• A. $2$
• B. $3$
• C. $4$
• D. $8$

Answer 1

The correct option is A $2$

$Formulaused$
${(a+b)}^3=a^3+b^3+3ab(a+b)$
$Given$
$x+\frac{1}{x}=2...............\left(i\right)$
$squaringbothsides$
$x^2+\frac{1}{x^2}=4-2$
$\Rightarrow x^2+\frac{1}{x^2}=2$
$squaringbothsides$
$x^4+\frac{1}{x^4}=4-2$
$\Rightarrow x^4+\frac{1}{x^4}=2$
$\Rightarrow (x^4+\frac{1}{x^4})(x+\frac{1}{x})=2(x+\frac{1}{x})$
$\Rightarrow x^5+\frac{1}{x^5}+x^3+\frac{1}{x^3}=2\left(2\right)from\left(i\right)$
$\Rightarrow x^5+\frac{1}{x^5}+\left(A\right)=4$
$\Rightarrow x^5+\frac{1}{x^5}=4-A.......\left(ii\right)$
$x+\frac{1}{x}=2$
$cubingbothsides$
${(x+\frac{1}{x})}^3=8$
$\Rightarrow x^3+\frac{1}{x^3}+3(x+\frac{1}{x})=8$
$\Rightarrow x^3+\frac{1}{x^3}=8-6$
$\Rightarrow x^3+\frac{1}{x^3}=2$
$pluggingin\left(ii\right)$
$x^5+\frac{1}{x^5}=4-2$
$\Rightarrow x^5+\frac{1}{x^5}=2$