Question

What is the sum of the squares of the following numbers?
$\frac{\sqrt{3}}{\sqrt{2}+1},\frac{\sqrt{3}}{\sqrt{2}-1},\frac{\sqrt{2}}{\sqrt{3}}$

• A. $16$
• B. $16\frac{2}{3}$
• C. $18$
• D. $18\frac{2}{3}$

Answer 1

Answer: (D)

$18\frac{2}{3}$

Required sum

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

$=\frac{3}{(\sqrt{2}+1{)}^2}+\frac{3}{(\sqrt{2}-1{)}^2}+\frac{2}{3}$

$=\frac{3}{2+1+2\sqrt{2}}+\frac{3}{2+1-2\sqrt{2}}+\frac{2}{3}$

$=\frac{3}{3+2\sqrt{2}}+\frac{3}{3-2\sqrt{2}}+\frac{2}{3}$

$=\frac{9(3-2\sqrt{2})+9(3+2\sqrt{2})+2(3+2\sqrt{2})(3-2\sqrt{2})}{3(3+2\sqrt{2})(3-2\sqrt{2})}$

$=\frac{27-18\sqrt{2}+27+18\sqrt{2}+2(9-8)}{3(9-8)}$

$=\frac{54+2}{3}=\frac{56}{3}=18\frac{2}{3}$