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If $(x-i y)^{1 / 3}-a-i b$, then the value of $\frac{x}{2 a}+\frac{y}{2 b}$ is
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Verified Answer
The correct answer is:
$2\left(a^2-b^2\right)$
$(x-i y)^{1 / 3}=a-i b$
$(x-i y)=(a-i b)^3$
$\begin{aligned} & =a^3-i^3 b^3-3 a^2(\mathrm{ib})+3 \mathrm{a}(\mathrm{ib})^2 \\ & =a^3+i b^3-3 a^2 b i-3 a b^2\end{aligned}$
$\begin{aligned} & x-i y=\left(a^3-3 a b^2\right)+i b\left(b^2-3 a^2\right) \\ & x=3 a^3-3 a b^2\end{aligned}$
$\frac{x}{a}=a^2-3 b^2$
$\frac{x}{2 a}=\frac{a^2-3 b^2}{2}$ ...(i)
and $y=b\left(3 a^2-b^2\right)$
$\frac{y}{2 b}=\frac{3 a^2-b^2}{2}$ ...(ii)
$y=b\left(3 a^2-b^2\right)$
(i) + (ii)
$\frac{x}{2 a}+\frac{y}{2 b}=\frac{a^2-3 b^2+3 a^2-b^2}{2}$
$\begin{aligned} & =\frac{4 a^2-4 b^2}{2} \\ & =2\left(a^2-b^2\right)\end{aligned}$
$(x-i y)=(a-i b)^3$
$\begin{aligned} & =a^3-i^3 b^3-3 a^2(\mathrm{ib})+3 \mathrm{a}(\mathrm{ib})^2 \\ & =a^3+i b^3-3 a^2 b i-3 a b^2\end{aligned}$
$\begin{aligned} & x-i y=\left(a^3-3 a b^2\right)+i b\left(b^2-3 a^2\right) \\ & x=3 a^3-3 a b^2\end{aligned}$
$\frac{x}{a}=a^2-3 b^2$
$\frac{x}{2 a}=\frac{a^2-3 b^2}{2}$ ...(i)
and $y=b\left(3 a^2-b^2\right)$
$\frac{y}{2 b}=\frac{3 a^2-b^2}{2}$ ...(ii)
$y=b\left(3 a^2-b^2\right)$
(i) + (ii)
$\frac{x}{2 a}+\frac{y}{2 b}=\frac{a^2-3 b^2+3 a^2-b^2}{2}$
$\begin{aligned} & =\frac{4 a^2-4 b^2}{2} \\ & =2\left(a^2-b^2\right)\end{aligned}$
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