Answer
$-32+32\sqrt3i$
Work Step by Step
RECALL:
De Moivre's Theorem: $(\cos{x}+i\sin{x})^n=\cos{(nx)}+i\sin{(nx)}$
Use the theorem above to obtain:
$[4(\cos40^\circ+i\sin40^\circ]^3
\\=4^3\left(\cos(3\cdot40^\circ)+i\cdot \sin(3\cdot40^\circ)\right)
\\=64\cdot(\cos(120^\circ)+i(\sin(120^\circ))
\\=64\left(-\frac{1}{2}+i\frac{\sqrt3}{2}\right)
\\=-32+32\sqrt3i$