Answer
At $x=3$, $y=108$ local maximum.
At $x=5$, $y=0$ local minimum.
Work Step by Step
Step 1. Given the function $y=x^3(x-5)^2$, take its derivative to get $y′=3x^2(x-5)^2+2x^3(x-5)=x^2(x-5)(3x-15+2x)=x^2(x-5)(3x-15+2x)==5x^2(x-5)(x-3)$
Step 2. The critical points can be found when $y'=0$ or undefined; we have $y′=5x^2(x-5)(x-3)=0$ which gives $x=0, 3, 5$
Step 3. At $x=0$, $y=0$ and we can identify this is not a local extrema (use local test points) as shown in the figure. Instead, this is an inflection point.
Step 4. At $x=3$, $y=3^3(3-5)^2=108$ and we can identify this as a local maximum (use local test points) as shown in the figure.
Step 5. At $x=5$, $y=5^3(5-5)^2=0$ and we can identify this as a local minimum (use local test points) as shown in the figure.