Question

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

• A. $27x^3+\frac{1}{27x^3}=956$
• B. $27x^3+\frac{1}{27x^3}=212$
• C. $27x^3+\frac{1}{27x^3}=488$
• D. $27x^3+\frac{1}{27x^3}=567$

Answer 1

The correct option is C $27x^3+\frac{1}{27x^3}=488$

$3x+\frac{1}{3x}=8$

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,

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

$=>27x^3+\frac{1}{27x^3}=8(62-1)$

$=8\left(61\right)$

$=488$