Question

If $x+\frac{1}{x}=4then{x}^5+\frac{1}{x^5}$ is equal to

• A. $723$
• B. $725$
• C. $756$
• D. $724$

Answer 1

Answer: (D)

$724$

$x+\frac{1}{x}=4$

squaring both sides

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

$\Rightarrow x^2+\frac{1}{x^2}+2x.\frac{1}{x}=16$

$\Rightarrow x^2+\frac{1}{x^2}=16-2=\mathrm{14....}\left(1\right)$

again cubing both sides

${(x+\frac{1}{x})}^3=4^3$

$\Rightarrow x^3+\frac{1}{x^3}+3x.\frac{1}{x}(x+\frac{1}{x})=64$

$\Rightarrow x^3+\frac{1}{x^3}+3\times 4=64$

$\Rightarrow x^3+\frac{1}{x^3}=64-12=\mathrm{52......}\left(2\right)$

Now multiplying (1) and (2) we get

$\Rightarrow (x^2+\frac{1}{x^2})(x^3+\frac{1}{x^3})=14\times 52$

$\Rightarrow x^5+\frac{x^2}{x^3}+\frac{x^3}{x^2}+\frac{1}{x^5}=14\times 52$

$\Rightarrow x^5+\frac{1}{x^5}+(\frac{1}{x}+x)=728$

$\Rightarrow x^5+\frac{1}{x^5}=728-4\Rightarrow x^5+\frac{1}{x^5}=724$