$$r^3-r^3-32r+60 = 0$$
Answer
$$ r = \frac{15}{8} $$
Explanation
$$ \begin{aligned} r^3-r^3-32r+60 &= 0&& \text{simplify left side} \\[1 em]r^3-r^3-32r+60 &= 0&& \\[1 em]-32r+60 &= 0&& \text{ move the constants to the right } \\[1 em]-32r &= -60&& \text{ divide both sides by $ -32 $ } \\[1 em]r &= \frac{-60}{-32}&& \\[1 em]r &= \frac{15}{8}&& \\[1 em] \end{aligned} $$
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