Question

Find the value of a and b

Answer 1

Answer:

$\large\boxed{a=\dfrac{58}{25},\ b=0}$

Step-by-step explanation:

$\dfrac{3\sqrt3+\sqrt2}{3\sqrt3-\sqrt2}+\dfrac{3\sqrt3-\sqrt2}{3\sqrt3+\sqrt2}\qquad\text{use}\ (a-b)(a+b)=a^2-b^2\\\\=\dfrac{(3\sqrt3+\sqrt2)(3\sqrt3+\sqrt2)}{(3\sqrt3-\sqrt2)(3\sqrt3+\sqrt2)}+\dfrac{(3\sqrt3-\sqrt2)(3\sqrt3-\sqrt2)}{(3\sqrt3+\sqrt2)(3\sqrt3-\sqrt2)}\\\\=\dfrac{(3\sqrt3+\sqrt2)^2}{(3\sqrt3)^2-(\sqrt2)^2}+\dfrac{(3\sqrt3-\sqrt2)^2}{(3\sqrt3)^2-(\sqrt2)^2}\qquad\text{use}\ (a\pm b)^2=a^2\pm2ab+b^2$

$=\dfrac{(3\sqrt3)^2+2(3\sqrt3)(\sqrt2)+(\sqrt2)^2}{3^2(\sqrt3)^2-2}+\dfrac{(3\sqrt3)^2-2(3\sqrt3)(\sqrt2)+(\sqrt2)^2}{3^2(\sqrt3)^2-2}\\\\=\dfrac{3^2(\sqrt3)^2+6\sqrt6+2}{9\cdot3-2}+\dfrac{3^2(\sqrt3)^2-6\sqrt6+2}{9\cdot3-2}\\\\=\dfrac{9\cdot3+6\sqrt6+2}{27-2}+\dfrac{9\cdot3-6\sqrt6+2}{27-2}\\\\=\dfrac{27+6\sqrt6+2}{25}+\dfrac{27-6\sqrt6+2}{25}\\\\=\dfrac{29+6\sqrt6}{25}+\dfrac{29-6\sqrt6}{25}\\\\=\dfrac{29+6\sqrt6+29-6\sqrt6}{25}\\\\=\dfrac{58}{25}$