Question

If $x^4+\frac{1}{x^4}=119$ and $x>1$, then the value of $x^3-\frac{1}{x^3}$ is

• A. $54$
• B. $18$
• C. $72$
• D. $36$

Answer 1

Answer: (D)

$36$

Given $x^4+\frac{1}{x^4}=119$

To find $x^3-\frac{1}{x^3}$

Sol: $x^4+\frac{1}{x^4}=119$

adding 2 both sides we get

$x^4+\frac{1}{x^4}+2=119+2=121$

${(x^2+\frac{1}{x^2})}^2={11}^2$ ............... Using $(a+b{)}^2=a^2+2ab+b^2$

$x^2+\frac{1}{x^2}=11$

Subtracting 2 from both sides, we get

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

${(x-\frac{1}{x})}^2=3^2$ ................ Using $(a-b{)}^2=a^2-2ab+b^2$

$x-\frac{1}{x}=3$

Cubing both sides we get,

$x^3-\frac{1}{x^3}-3(x-\frac{1}{x})=27$ ................... Using $(a-b{)}^3=a^3-b^3-3ab(a-b)$

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

$x^3-\frac{1}{x^3}=27+9=36$

$\therefore x^3-\frac{1}{x^3}=36$