Solving for A

$(4-a)^3\ =\ 32$

                                         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.

$4-2\sqrt[3]{4}\ =\ a$

$a\ =\ 4-2\sqrt[3]{4}$-

(4-a)^3 = 32      cube root both sides   (note that 32 = 2^5 )

(4-a) = cubrt(2^5)      Simplify the right side

(4-a) = 2 cubrt(4)      Add 'a' to both sides and subtract 2 cubrt(4) from both sides

4- 2 cubrt(4) = a