Question

(y^2+6y-17)^((1)/(3))-(y^2+3y-5)^((1)/(3))=0

Answer 1

\[\frac{\sqrt[3]{y^2 + 6y - 17}}{\sqrt[3]{y^2 + 3y - 5}} = 0\]
\[\sqrt[3]\frac{{y^2 + 6y - 17}}{{y^2 + 3y - 5}} = 0\]

\[\left(\sqrt[3]\frac{{y^2 + 6y - 17}}{{y^2 + 3y - 5}}\right)^3 = 0\]
\[\frac{{y^2 + 6y - 17}}{{y^2 + 3y - 5}}= 0\]

\[y^2 + 6y - 17 = 0\]

Finish solving by completing the square.

Answer 2

@salce08