Question

27y^3+125=0 find y, imaginary and real numbers

Answer 1

\(27y^3 = (3y)^3\)
\(125 = 5^3\)
So that your equation is the form of \(a^3 +b^3=(a+b)(a^2-ab+b^2)=0\) where a = 3y, b = 5
Expand it and solve for y