Question

If $x=3-\sqrt{8}$, then $x^3+\frac{1}{x^3}$ is equal to ?

• A. $6$
• B. $198$
• C. $6\sqrt{2}$
• D. $102$

Answer 1

Answer: (B)

$198$

$x=3-\sqrt{8}$
$\frac{1}{x}=\frac{1}{3-\sqrt{8}}=\frac{3+\sqrt{8}}{3^2-{\sqrt{8}}^2}$
$\frac{1}{x}=3+\sqrt{8}$
$⟹x+\frac{1}{x}=6$
Cubing both sides:
$x^3+\frac{1}{x^3}+3(x\cdot \frac{1}{x})(x+\frac{1}{x})=216$
$x^3+\frac{1}{x^3}+3\cdot 1\cdot 6=216$
$⟹x^3+\frac{1}{x^3}=198$