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The cube roots of $-1$ are?

$\begin{array}{1 1} -1,\large\frac{1}{2}(-1\pm i\sqrt 3) \\ -1,\large\frac{1}{2}(1\pm i\sqrt 3) \\ 1,\large\frac{1}{2}(-1\pm i\sqrt 3) \\ -1,-1,-1 \end{array}$

1 Answer

Let $(-1)^{\large\frac{1}{3}}=x$
$\Rightarrow\:x^3=-1$
$\Rightarrow\:x^3+1=0$
$\Rightarrow\:(x+1)(x^2-x+1)=0$
$\Rightarrow\:x=-1,\large\frac{1}{2}$$(1\pm \sqrt 3 i)$
answered Aug 1, 2013 by rvidyagovindarajan_1
 

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