Question

Find all solutions of the equation. leave answers in trigonometric form x^3-8=0 2cis

Answer 1

$x^3-8=0$

$x^3=8=8\mathrm{cis}\,0^\circ$

By DeMoivre's theorem, we have

$x=8^{1/3}\mathrm{cis}\left(\dfrac{0^\circ+360^\circ k}3\right)$

with $k=0,1,2$. Then we have three solutions,

$x=2\mathrm{cis}\,0^\circ=2$

$x=2\mathrm{cis}\,120^\circ$

$x=2\mathrm{cis}\,240^\circ$