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One of the values of $(i)^{\large\frac{1}{3}}$ is

(A) $\large\frac{-1+\sqrt 3 i}{2}$
(B) $\large\frac{-1-\sqrt 3 i}{2}$
(C) -i
(D) i
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1 Answer

Toolbox:
  • $i^3=-i$
  • $a^3+b^3=(a+b)(a^2-ab+b^2)$
Let $(i)^{\large\frac{1}{3}}$$=x$
cubing both the sides we get
$i=x^3$
$\Rightarrow\:x^3-i=0$
$\Rightarrow\:x^3+i^3=0$
$\Rightarrow\:(x+i)(x^2-ix-1)$
$\Rightarrow\:x=-i,\:\:or\:\:x=\large\frac{i\pm \sqrt {3}}{2}$
answered Jul 21, 2013 by rvidyagovindarajan_1
 

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