Question

If $x={\left(\frac{a}{b}\right)}^{\frac{2ab}{a^2-b^2}}$, shew that $\frac{ab}{a^2+b^2}(x^{\frac{a}{b}}+x^{\frac{b}{a}})={\left(\frac{a}{b}\right)}^{\frac{a^2+b^2}{a^2-b^2}}$

Answer 1

Given,

$x=(\frac{a}{b}{)}^{\frac{2ab}{a^2-b^2}}$

$(x{)}^{\frac{a}{b}}=(\frac{a}{b}{)}^{\frac{2a^2}{a^2-b^2}}$

$(x{)}^{\frac{b}{a}}=(\frac{a}{b}{)}^{\frac{2b^2}{a^2-b^2}}$

Taking L.H.S.

$\frac{ab}{a^2+b^2}(x^{\frac{a}{b}}+x^{\frac{b}{a}})$

$\to \frac{ab}{a^2+b^2}\left(\right(\frac{a}{b}{)}^{\frac{2a^2}{a^2-b^2}}+\left(\frac{a}{b}{)}^{\frac{2b^2}{a^2-b^2}}\right)$

$\to \frac{ab}{a^2+b^2}(\frac{a^{\frac{2a^2}{a^{-}{b}^2}}}{b^{\frac{2a^2}{a^2-b^2}}}+\frac{a^{\frac{2b^2}{a^2-b^2}}}{b^{\frac{2b^2}{a^2-b^2}}})$

Taking L.C.M

$\to \frac{ab}{a^2+b^2}\left(\frac{a^{\frac{2a^2}{a^2-b^2}}.b^{\frac{2b^2}{a^2-b^2}}+a^{\frac{2b^2}{a^2-b^2}}.b^{\frac{2a^2}{a^2-b^2}}}{b^{\frac{2a^2+2b^2}{a^2-b^2}}}\right)$

Multiplying ab inside

$\to \frac{1}{a^2+b^2}\left(\frac{a^{\frac{3a^2-b^2}{a^2-b^2}}.b^{\frac{a^2+b^2}{a^2-b^2}}+a^{\frac{a^2+b^2}{a^2-b^2}}.b^{\frac{3a^2-b^2}{a^2-b^2}}}{b^{\frac{2a^2+2b^2}{a^2-b^2}}}\right)$

Taking common

$\to \frac{a^{\frac{a^2+b^2}{a^2-b^2}}.b^{\frac{a^2+b^2}{a^2-b^2}}}{a^2+b^2}\left(\frac{a^{\frac{2a^2-2b^2}{a^2-b^2}}+b^{\frac{2a^2-2b^2}{a^2-b^2}}}{b^{\frac{2a^2+2b^2}{a^2-b^2}}}\right)$

$\to \frac{a^{\frac{a^2+b^2}{a^2-b^2}}}{a^2+b^2}\left(\frac{a^2+b^2}{b^{\frac{a^2+b^2}{a^2-b^2}}}\right)$

$\to (\frac{a}{b}{)}^{\frac{a^2+b^2}{a^2-b^2}}$

=R.H.S