Let left(-2-frac{1}{3} iright)^3=frac{x+i y}{ | vedclass

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(D) Given $\left(-2-\frac{1}{3} i\right)^3=\frac{x+i y}{27}$.We can rewrite the left side as $\left(\frac{-6-i}{3}\right)^3 = \frac{(-6-i)^3}{27}$.Equ

Vedclass · Accepted Answer

$91$

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(D) Given $\left(-2-\frac{1}{3} i\right)^3=\frac{x+i y}{27}$.We can rewrite the left side as $\left(\frac{-6-i}{3}\right)^3 = \frac{(-6-i)^3}{27}$.Equating the numerators,we have $x+iy = (-6-i)^3$.Using the identity $(a+b)^3 = a^3+3a^2b+3ab^2+b^3$:$(-6-i)^3 = (-6)^3 + 3(-6)^2(-i) + 3(-6)(-i)^2 + (-i)^3$.$= -216 + 3(36)(-i) + 3(-6)(-1) - (-i)$.$= -216 - 108i + 18 + i$.$= -198 - 107i$.Comparing with $x+iy$,we get $x = -198$ and $y = -107$.Therefore,$y-x = -107 - (-198) = -107 + 198 = 91$.

Let $\left(-2-\frac{1}{3} i\right)^3=\frac{x+i y}{27}$,where $i=\sqrt{-1}$ and $x, y$ are real numbers. Then the value of $(y-x)$ is:

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