# show that

Question:

If $a=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}$ and $b=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}$, show that $3 a^{2}+4 a b-3 b^{2}=4+\frac{56}{3} \sqrt{10}$.

Solution:

According to question,

$a=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}$ and $b=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}$

$a=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}$

$=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}} \times \frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}}$

$=\frac{(\sqrt{5}+\sqrt{2})^{2}}{(\sqrt{5})^{2}-(\sqrt{2})^{2}}$

$=\frac{(\sqrt{5})^{2}+(\sqrt{2})^{2}+2 \sqrt{5} \sqrt{2}}{5-2}$

$=\frac{5+2+2 \sqrt{10}}{3}$

$=\frac{7+2 \sqrt{10}}{3} \quad \ldots(1)$

$b=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}$

$=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}} \times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}$

$=\frac{(\sqrt{5}-\sqrt{2})^{2}}{(\sqrt{5})^{2}-(\sqrt{2})^{2}}$

$=\frac{(\sqrt{5})^{2}+(\sqrt{2})^{2}-2 \sqrt{5} \sqrt{2}}{5-2}$

$=\frac{5+2-2 \sqrt{10}}{3}$

$=\frac{7-2 \sqrt{10}}{3} \ldots(2)$

Now,

$3 a^{2}+4 a b-3 b^{2}$

$=3\left(a^{2}-b^{2}\right)+4 a b$

$=3(a+b)(a-b)+4 a b$

$=3\left(\frac{7+2 \sqrt{10}}{3}+\frac{7-2 \sqrt{10}}{3}\right)\left(\frac{7+2 \sqrt{10}}{3}-\frac{7-2 \sqrt{10}}{3}\right)+4\left(\frac{7+2 \sqrt{10}}{3} \times \frac{7-2 \sqrt{10}}{3}\right)$

$=3\left(\frac{14}{3}\right)\left(\frac{4 \sqrt{10}}{3}\right)+4\left(\frac{(7)^{2}-(2 \sqrt{10})^{2}}{9}\right)$

$=\frac{56}{3} \sqrt{10}+4\left(\frac{49-40}{9}\right)$

$=\frac{56}{3} \sqrt{10}+4$

Hence, $3 a^{2}+4 a b-3 b^{2}=4+\frac{56 \sqrt{10}}{3}$