Let \( z = y^{\frac{1}{3}} \). Then \( z^2 = y^{\frac{2}{3}} \), and the equation is:
\[\nz^2 - 2z = 15\n\]
Rearranging:
\[\nz^2 - 2z - 15 = 0\n\]
Factoring:
\[\n(z - 5)(z + 3) = 0\n\]
Thus, \( z = 5 \) or \( z = -3 \). Since \( z = y^{\frac{1}{3}} \):
\[\ny^{\frac{1}{3}} = 5 \quad \Rightarrow \quad y = 25\n\]
\[\ny^{\frac{1}{3}} = -3 \quad \Rightarrow \quad y = -27\n\]
Therefore, the solutions are \( y = 25 \) and \( y = -27 \).