Question

If $x=\frac{1}{5-x}$ and $x\ne 5$, find the value of

$x^3+\frac{1}{x^3}$

• A. $80$
• B. $240$
• C. $110$
• D. $530$

Answer 1

The correct option is C $110$

$x=\frac{1}{5-x}$
$=>5-x=\frac{1}{x}$
$=>x+\frac{1}{x}=5$
Now,
Squaring both sides
$(x+\frac{1}{x}{)}^2=(5{)}^2$
$=>x^2+\frac{1}{x^2}+2\left(x\right)\left(\frac{1}{x}\right)=25$
$=>x^2+\frac{1}{x^2}=25-2$
$=>x^2+\frac{1}{x^2}=23$
Now,
$(x^3+\frac{1}{x^3})=(x+\frac{1}{x})\left(x^2\right)+\left(\frac{1}{x^2}\right)-\left(x\right)\left(\frac{1}{x}\right)$
$=\left(5\right)(23-1)$
$=110$