Question

If $x=3-2\sqrt{2}$, find the value of $x^3+\frac{1}{x^3}$.

Answer 1

It is given that $x=3-2\sqrt{2}$, so we first find $x^3$ that is ${(3-2\sqrt{2})}^3$ as follows:

${(3-2\sqrt{2})}^3=(3-2\sqrt{2})\{{\left(3\right)}^2+{\left(2\sqrt{2}\right)}^2-(2\times 3\times 2\sqrt{2})\}\{\because {(a-b)}^3=(a-b)({a^2+b}^2-2ab)\}=(3-2\sqrt{2})\{9+8-\left(12\sqrt{2}\right)\}=(3-2\sqrt{2})(17-12\sqrt{2})$

$=(3\times 17)+[3\times (-12\sqrt{2})]+[(-2\sqrt{2})\times 17]+[(-2\sqrt{2})\times (-12\sqrt{2})]=51-36\sqrt{2}-34\sqrt{2}+48=99-70\sqrt{2}$

Therefore, $x^3=99-70\sqrt{2}$

Now we will find the value of $x^3+\frac{1}{x^3}$ as shown below:

$x^3+\frac{1}{x^3}=99-70\sqrt{2}+\frac{1}{99-70\sqrt{2}}=99-70\sqrt{2}+(\frac{1}{99-70\sqrt{2}}\times \frac{99+70\sqrt{2}}{99+70\sqrt{2}})\left\{Rationalizedenominator\right\}=99-70\sqrt{2}+\left(\frac{99+70\sqrt{2}}{(99-70\sqrt{2})(99+70\sqrt{2})}\right)=99-70\sqrt{2}+\left(\frac{99+70\sqrt{2}}{{\left(99\right)}^2-{\left(70\sqrt{2}\right)}^2}\right)\{\because (a-b)(a+b)=a^2-b^2\}$

$=99-70\sqrt{2}+\left(\frac{99+70\sqrt{2}}{9801-9800}\right)=99-70\sqrt{2}+\left(\frac{99+70\sqrt{2}}{1}\right)=99-70\sqrt{2}+99+70\sqrt{2}=99+99=198$

Hence, $x^3+\frac{1}{x^3}=198$