How do you find the cube roots of $- 125$ $-125$ ?

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How do you find the cube roots of $- 125$ $-125$ ?

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Answer 1 · 2016-09-08T09:49:31

$5 (\frac{1}{2} + \frac{i \sqrt{3}}{2}), - 5, 5 (\frac{1}{2} - \frac{i \sqrt{3}}{2})$ $5(1/2 + (i sqrt[3])/2),-5,5(1/2 - (i sqrt[3])/2)$

Explanation:

Using the de Moivre's identity

$- 125 = 5^{3} e^{i (π + 2 k π)}, k = 0, \pm 1, \pm 2, \dots$ $-125 = 5^3 e^(i(pi+2kpi)), k=0,pm1,pm2,cdots$ because

$e^{i (π + 2 k π)} = cos (π + 2 k π) + i sin (π + 2 k π) = - 1$ $e^(i(pi+2kpi)) = cos(pi+2kpi)+i sin(pi+2kpi) = -1$

so

${(- 125)}^{\frac{1}{3}} = {(5^{3} e^{i (π + 2 k π)})}^{\frac{1}{3}} = 5 e^{i \frac{π + 2 k π}{3}}$ $(-125)^(1/3) = (5^3 e^(i(pi+2kpi)))^(1/3) = 5e^(i(pi+2kpi)/3)$

for $k = 0, k = \pm 1$ $k=0, k=pm1$ we have

$5 e^{i \frac{π}{3}}, 5 e^{i π}, 5 e^{- i \frac{π}{3}}$ $5e^(ipi/3),5e^(ipi),5e^(-ipi/3)$

or

$5 (\frac{1}{2} + \frac{i \sqrt{3}}{2}), - 5, 5 (\frac{1}{2} - \frac{i \sqrt{3}}{2})$ $5(1/2 + (i sqrt[3])/2),-5,5(1/2 - (i sqrt[3])/2)$

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How do you find the cube roots of $- 125$ $-125$ ?

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