If a and b are distinct positive primes such that

$\sqrt[3]{a^{6} b^{-4}}=a^{x} b^{2} y$

$\left(a^{6} b^{-4}\right)^{1 / 3}=a^{x} b^{2 y}$

$a^{6 / 3} b^{-4 / 3}=a^{x} b^{2 y}$

$a^{2} b^{-4 / 3}=a^{x} b^{2 y}$

x = 2, 2y = −4/3

$y=\frac{\frac{-4}{3}}{2}$

$y=-\frac{2}{3}$