Question

If $3x+\frac{1}{3x}=3$, then find $27x^3+\frac{1}{27x^3}$.

• A. 9
• B. 27
• C. 18
• D. 36

Answer 1

The correct option is C 18

Given, $3x+\frac{1}{3x}=3$
Cubing both sides,

${(3x+\frac{1}{3x})}^3=27$

Applying the formula,${(a+b)}^3=a^3+b^3+3ab(a+b)$

$(3x{)}^3+{\left(\frac{1}{3x}\right)}^3+3\times 3x\times \frac{1}{3x}\times (3x+\frac{1}{3x})=27$
$27x^3+\frac{1}{27x^3}+3\times 3=27$
$=>27x^3+\frac{1}{27x^3}=27-9=18$