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

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Aicharya · Accepted Answer

$-(6+i)^{3}=x+i y$ $\Rightarrow \quad-\left[216+i^{3}+18 i(6+i)\right]=x+i y$ $\Rightarrow \quad-[216-i+108 i-18]=x+i y$ $\Rightarrow \quad-216+i-108 i+18=x+i y$ $\Rightarrow-198-107 i=x+i y$ $\Rightarrow \quad x=-198, y=-107$ $\Rightarrow \quad y-x=-107+198=91$

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

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