Find $x+y$ given that $(x+4)^{50} + (y-5)^{100} = 0$

Question

Find $x+y$ given that $(x+4)^{50} + (y-5)^{100} = 0$.

I am stuck on this; any hint on the first step will be appreciated. Thought of taking the $50$th root on both sides, but I don't think that's the right step. Thank you.

Answer 1

If $a^2 = 0$, then $a = 0$.

If $a^2 + b^2 = 0$, then $a = b = 0$.

If $((x+4)^{25})^2 + ((y-5)^{50})^2 = 0$, then $(x+4)^{25} = (y-5)^{50} = 0$, i.e. $x+4 = y-5 = 0$, i.e. $x = -4$ and $y = 5$.