Question

If $x^2+\frac{1}{x^2}=7$ then the vaue of $x^3+\frac{1}{x^3}$ is

• A. $9$
• B. $18$
• C. $27$
• D. $14$

Answer 1

The correct option is B $18$

Given, $x^2+\frac{1}{x^2}=7$

We know, ${(x+\frac{1}{x})}^2=x^2+\frac{1}{x^2}+2$

${(x+\frac{1}{x})}^2=7+2=9$

$\therefore x+\frac{1}{x}=3$

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

$27=x^3+\frac{1}{x^3}+3\left(3\right)$

$\therefore x^3+\frac{1}{x^3}=27-9=18$