Question

$If{x}^4+\frac{1}{x^4}=119,thenthevalueof{x}^3-\frac{1}{x^3}$ is

• A. 18
• B. 24
• C. 27
• D. 36

Answer 1

The correct option is D 36

If $x^4-\frac{1}{x^4}=119$
$\Rightarrow (x^2+\frac{1}{x^2}{)}^2-2=119$
$\Rightarrow (x^2+\frac{1}{x^2}{)}^2=121$
$\Rightarrow x^2+\frac{1}{x^2}=11(Negativevalueisnotpossiblefor{x}^2+\frac{1}{x^2})$
We can write, $(x+\frac{1}{x}{)}^2-2=11$
$\Rightarrow (x+\frac{1}{x}{)}^2=13$
$\because x-\frac{1}{x}=\sqrt{(x+\frac{1}{x}{)}^2-4}$
=$\sqrt{13-4}=\sqrt{9}$
$x-\frac{1}{x}=3$
So, $x^3-\frac{1}{x^3}=(x-\frac{1}{x}{)}^3+3(x-\frac{1}{x})$
$=(3{)}^3+3\times 3$
=27+9
=36