Question

If $x^2=y^3$ then prove that ${\left(\frac{x}{y}\right)}^{\frac{3}{2}}+{\left(\frac{y}{x}\right)}^{\frac{2}{3}}=x^{\frac{1}{2}}+y^{-\frac{1}{3}}$

Answer 1

Given, $x^2=y^3$.......(1).

Now,

${\left(\frac{x}{y}\right)}^{\frac{3}{2}}+{\left(\frac{y}{x}\right)}^{\frac{2}{3}}$

$={\left(\frac{x^2}{y^2}\right)}^{\frac{3}{4}}+{\left(\frac{y^2}{x^2}\right)}^{\frac{1}{3}}$

$={\left(\frac{y^3}{y^2}\right)}^{\frac{3}{4}}+{\left(\frac{y^2}{y^3}\right)}^{\frac{1}{3}}$ [ Using (1)]

$=y^{\frac{3}{4}}+y^{-\frac{1}{3}}$

$=(y^3{)}^{\frac{1}{4}}+y^{-\frac{1}{3}}$

$=(x^2{)}^{\frac{1}{4}}+y^{-\frac{1}{3}}$ [ Using (1)]

$=(x{)}^{\frac{1}{2}}+y^{-\frac{1}{3}}$.