Let $\left(-2-\frac{1}{3}\right)^3=\frac{x+\mathrm{i} y}{27}, \mathrm{i}=\sqrt{-1}$, where $x$ and $y$ are real numbers, then $(y-x)$ has the value

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Aicharya · Accepted Answer

$\begin{aligned} & \left(-2-\frac{1}{3} \mathrm{i}\right)^3=\frac{x+\mathrm{i} y}{27} \\ & \left(-2-\frac{1}{3} \mathrm{i}\right)^3=\frac{1}{27}(-6-\mathrm{i})^3 \\ & \text { Consider, }(-6-\mathrm{i})^3 \\ & =(-6)^3+3(-6)^2(-\mathrm{i})+3(-6)(-\mathrm{i})^2+(-\mathrm{i})^3 \\ & =-216-108 \mathrm{i}+18+\mathrm{i} \\ & =-198-107 \mathrm{i} \\ \therefore \quad & \left(-2-\frac{1}{3} \mathrm{i}\right)^3=\frac{-198-107 \mathrm{i}}{27} \end{aligned}$
Comparing with $\frac{x+i y}{27}$, we get $\begin{aligned} & x=-198, y=-107 \\ & y-x=-107+198=91 \end{aligned}$

Let $\left(-2-\frac{1}{3}\right)^3=\frac{x+\mathrm{i} y}{27}, \mathrm{i}=\sqrt{-1}$, where $x$ and $y$ are…

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