Answer
$4\ in \times 4\ in \times 4\ in$.
Work Step by Step
Step 1. Use the figure and dimensions given in the exercise, we have the volume after cut off as
$V=x^2(x-2)=32$
Step 2. Use synthetic division or graphing to solve the above equation: $x^3-2x^2-32=0$ or $(x-4)(x^2+2+8)=0$, thus $x=4\ in$
Step 3. The dimensions of the cube are: $4\ in \times 4\ in \times 4\ in$.
