If $(x+i y)^{1 / 3}=a+i b$ where $x, y, a, b \in R$ and $i=\sqrt{-1}$, then $\frac{x}{a}-\frac{y}{b}=$

If $(x+i y)^{1 / 3}=a+i b$ where $x, y, a, b \in R$ and $i=\sqrt{-1}$, then $\frac{x}{a}-\frac{y}{b}=$
  1. $-2\left(a^2+b^2\right)$
  2. $2\left(a^2-b^2\right)$
  3. $a^2-b^2$
  4. $a^2+b^2$

Solution

$\begin{aligned} & (x+i y)^{1 / 3}=a+i b \Rightarrow x+i y=(a+i b)^3=\left(a^3-3 a b^2\right)+i\left(3 a^2 b-b^3\right) \\ & \Rightarrow \frac{x}{a}=a^2-3 b^2 \text { and } \frac{y}{b}=3 a^2-b^2 \\ & \Rightarrow \frac{x}{a}-\frac{y}{b}=-2\left(a^2+b^2\right)\end{aligned}$

Asked in: MHT CET 2022 (06 Aug Shift 1)

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