Question

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

Answer 1

It is given that $3x+\frac{1}{3x}=3$, by taking cubes on both sides we get:

${(3x+\frac{1}{3x})}^3=3^3\Rightarrow {\left(3x\right)}^3+{\left(\frac{1}{3x}\right)}^3+(3\times 3x\times \frac{1}{3x})(3x+\frac{1}{3x})=27(\because (a+b{)}^3=a^3+b^3+3ab(a+b))\Rightarrow 27x^3+\frac{1}{27x^3}+3\left(3\right)=27(Given3x+\frac{1}{3x}=3)\Rightarrow 27x^3+\frac{1}{27x^3}+(3\times 3)=27\Rightarrow 27x^3+\frac{1}{27x^3}+9=27\Rightarrow 27x^3+\frac{1}{27x^3}=27-9\Rightarrow 27x^3+\frac{1}{27x^3}=18$

Hence, $27x^3+\frac{1}{27x^3}=18$.