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

asked Jun 20, 2020 17.2k views

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x^3-8=0

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

By DeMoivre's theorem, we have

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

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

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

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

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

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