\((4-a)^3\ =\ 32\)
Take the cube root of both sides of the equation.
\(\sqrt[3]{(4-a)^3}\ =\ \sqrt[3]{32}\)
Simplify the left side with the rule \(\sqrt[3]{n^3}\ =\ n\)
\(4-a\ =\ \sqrt[3]{32}\)
We can rewrite 32 like this because 32 = 2 * 2 * 2 * 2 * 2
\(4-a\ =\ \sqrt[3]{2\cdot2\cdot2\cdot2\cdot2}\)
We can rewrite the right side again like this...
\(4-a\ =\ \sqrt[3]{2\cdot2\cdot2}\cdot\sqrt[3]{2\cdot2}\)
And 2 * 2 * 2 = 23 and 2 * 2 = 4
\(4-a\ =\ \sqrt[3]{2^3}\,\cdot\,\sqrt[3]{4}\)
Simplify \(\sqrt[3]{2^3}\) again with the rule \(\sqrt[3]{n^3}\ =\ n\)
\(4-a\ =\ 2\,\cdot\,\sqrt[3]{4}\)
\(4-a\ =\ 2\sqrt[3]{4}\)
Add a to both sides of the equation.
\(4\ =\ 2\sqrt[3]{4}+a\)
Subtract \(2\sqrt[3]{4}\) from both sides of the equation.
\(4-2\sqrt[3]{4}\ =\ a\)
\(a\ =\ 4-2\sqrt[3]{4}\)-
So, unless the criminal car is very far away, it won’t take long to catch up to it. 
