If $y^2= 36$, what is the greatest possible value of $y^3$?
I notice that the equation $y^2 = 36$ has two possible solutions: $y = \pm 6$. To find the greatest possible value of $y^3$, I need to compare the values of $6^3$ and $(-6)^3$. I recall that raising a negative number to an odd power preserves the negative sign, so $(-6)^3 = -216$. I also recall that raising a positive number to any power keeps it positive, so $6^3 = 216$. Comparing $-216$ and $216$, I see that $216$ is greater. Therefore, the greatest possible value of $y^3$ is $216$. 