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# If $\;(x+iy)^{\large\frac{1}{3}}= a+ib\;$ where $\;x\;,y\;,a\;,b \in R\;$ show that $\; \large\frac{x}{a} - \large\frac{y}{b} = -2 (a^{2} + b^{2})$

$(a)\;-2(a^{2}+b^{2})\qquad(b)\;2(a^{2}-b^{2})\qquad(c)\;a^{2}+b^{2}\qquad(d)\;2(b^{2}-a^{2})$

Answer : $\;-2(a^{2}+b^{2})$
Explanation :
$\;(x+iy)^{\large\frac{1}{3}}= a+ib\;$
$\;(x+iy)^{\large\frac{1}{3}}= (a+ib)^{3}\;$
$= a^{3}+(ib)^{3} + 3\;.a\;.ib\;(a+ib)$
$= a^{3} -ib^{3} +3a^{2}b i +3ab^{2}i^{2}$
$= a^{3} -ib^{3} +3a^{2}b i-3ab^{2}$
$= (a^{3}-3ab^{2})+ i (3a^{2}b -b^{3})$
$x = (a^{3}-3ab^{2})\;$ and $\;y=(3a^{2}b -b^{3})$
$\; \large\frac{x}{a} - \large\frac{y}{b} = \large\frac{a^{3} - 3ab^{2}}{a} - \large\frac{3a^{2}b-b^{3}}{b}$
$= \large\frac{a(a^{2} - 3b^{2})}{a} - \large\frac{b(3a^{2}-b^{2})}{b}$
$= a^{2} -3b^{2} - 3a^{2} +b^{2}$
$= -2a^{2} -2b^{2}$
$=-2(a^{2}+b^{2})\;.$