Question

If $x\ne 0$ and $3x+\frac{1}{3x}=8$, find the value of :
$27x^3+\frac{1}{27x^3}$

• A. $642$
• B. $462$
• C. $169$
• D. $488$

Answer 1

The correct option is D $488$

$3x+\frac{1}{3x}=8$
Squaring both sides,
$=>(3x+(\frac{1}{3x}{)}^2=8^2$
$=>(3x{)}^2+(\frac{1}{3x}{)}^2+2\left(3x\right)\left(\frac{1}{3x}\right)=64$
$=>(3x{)}^2+(\frac{1}{3x}{)}^2+2=64$
$=>(3x{)}^2+(\frac{1}{3x}{)}^2=64-2$
$=>(3x{)}^2+(\frac{1}{3x}{)}^2=62$
Now,
$=(3x{)}^3+\frac{1}{3x^3}=(3x+\frac{1}{3x}\left)\right[\left(3x^2\right)-\left(3x\right)\left(\frac{1}{3x}\right)+\left(\frac{1}{3x^2}\right)]$
$=>27x^3+\frac{1}{27x^3}=8(62-1)$
$=8\left(61\right)$
$=488$