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If $(x+iy)^{\large\frac{1}{3}}=a+ib$ then $\large\frac{x}{a}+\frac{y}{b}$ equals

$\begin{array}{1 1}(A)\;2(a^2+b^2)&(B)\;2(a^2-b^2)\\(C)\;4(a^2+b^2)&(D)\;4(a^2-b^2)\end{array} $

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$(x+iy)^{\large\frac{1}{3}}=a+ib$
$\Rightarrow x+iy=(a+ib)^3=a^3+3a^2(ib)+3a(ib)^2+(ib)^3$
$\Rightarrow x+iy=(a^3-3ab^2)+(3a^2b-b^3)i$
$\Rightarrow x=a^3-3ab^2$,$y=3a^2b-b^3$
$\Rightarrow \large\frac{x}{a}$$=a^2-3ab^2$,$\large\frac{y}{b}$$=3a^2-b^2$
$\large\frac{x}{a}+\frac{y}{b}$$=a^2-3b^2+3a^2-b^2$
$\Rightarrow 4(a^2-b^2)$
Hence (D) is the correct answer.
answered Apr 10, 2014 by sreemathi.v
 

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