Question

If $x^4+\frac{1}{x^4}=47$, then the value of $x^3+\frac{1}{x^3}$ is ......

• A. $18$
• B. $27$
• C. $14$
• D. $12$

Answer 1

The correct option is A $18$
Given,
$x^4+\frac{1}{x^4}=47$
We know,
${(x^2+\frac{1}{x^2})}^2=(x^2{)}^2+(\frac{1}{x^2}{)}^2+2\left(x^2\right)\left(\frac{1}{x^2}\right)$
$=x^4+\frac{1}{x^4}+2$
$=47+2$
$=49$
$=>x^2+\frac{1}{x^2}=7$
$=>(x+\frac{1}{x}{)}^2=x^2+\frac{1}{x^2}+2(x\left)\right(\frac{1}{x})$
$=7+2$
$=9$
$=>x+\frac{1}{x}=3$
We know that,
$x^3+\frac{1}{x^3}=(x+\frac{1}{x})(x^2+\frac{1}{x^2}-x\times \frac{1}{x})$
$=3(7-1)$
$=18$
Ans A