Question

If $x=\frac{1}{5-x}$ and $x\ne 5$, find : $x^3+\frac{1}{x^3}$.

• A. $95$
• B. $132$
• C. $110$
• D. $140$

Answer 1

Answer: (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,

$(x^3+\frac{1}{x^3})=(x+\frac{1}{x})\left[\right(x^2)+(\frac{1}{x^2})-(x\left)\right(\frac{1}{x})$ ]

$=\left(5\right)(23-1)$

$=110$