Question

If $m^2-5m+1=0$ , then the minimum value of $(m^3+\frac{1}{m^3})$ is ?

• A. $100$
• B. $105$
• C. $110$
• D. $130$

Answer 1

The correct option is C $110$

$m^2-5m+1=0$
$m^2+1=5m$
$m+\frac{1}{m}=5$
Cubing on both sides
$m^3+\frac{1}{m^3}+3m.\frac{1}{m}(m+\frac{1}{m})=5^3$
$m^3+\frac{1}{m^3}+3\left(5\right)=125$
$m^3+\frac{1}{m^3}=125-15$
$m^3+\frac{1}{m^3}=110$