Question

If $a=\frac{1}{3-2\sqrt{2}}$, $b=\frac{1}{3+2\sqrt{2}}$, then $a^3+b^3=$?

• A. $194$
• B. $196$
• C. $198$
• D. $200$

Answer 1

The correct option is C $198$

We know that one of the identity is $a^3+b^3=(a+b)(a^2+b^2-ab)$.

Here, $a=\frac{1}{3-2\sqrt{2}}$ and $b=\frac{1}{3+2\sqrt{2}}$. Now applying the identity we get:

$a^3+b^3=(\frac{1}{3-2\sqrt{2}}+\frac{1}{3+2\sqrt{2}})[{\left(\frac{1}{3-2\sqrt{2}}\right)}^2+{\left(\frac{1}{3+2\sqrt{2}}\right)}^2-(\frac{1}{3-2\sqrt{2}}\times \frac{1}{3+2\sqrt{2}})]=\left(\frac{3+2\sqrt{2}+3-2\sqrt{2}}{3^2-(2\sqrt{2}{)}^2}\right)[\left(\frac{1}{3^2+(2\sqrt{2}{)}^2-(2\times 3\times 2\sqrt{2})}\right)+\left(\frac{1}{3^2+(2\sqrt{2}{)}^2-(2\times 3\times 2\sqrt{2})}\right)-\left(\frac{1}{3^2-(2\sqrt{2}{)}^2}\right)](\because x^2-y^2=(x+y\left)\right(x-y),(x+y{)}^2=x^2+y^2+2xy,(x-y{)}^2=x^2+y^2-2xy)=\left(\frac{6}{9-8}\right)[\frac{1}{9+8-12\sqrt{2}}+\frac{1}{9+8+12\sqrt{2}}-\left(\frac{1}{9-8}\right)]=6[(\frac{1}{17-12\sqrt{2}}+\frac{1}{17+12\sqrt{2}})-1]=6[\left(\frac{17+12\sqrt{2}+17-12\sqrt{2}}{{17}^2-(12\sqrt{2}{)}^2}\right)-1]=6[\left(\frac{34}{289-288}\right)-1]=6(34-1)=6\times 33=198$

Hence, $a^3+b^3=198$ if $a=\frac{1}{3-2\sqrt{2}}$ and $b=\frac{1}{3+2\sqrt{2}}$.