Let {left( { - 2 - frac{1}{3}i} right)^3} = f | vedclass

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(A) Given: ${\left( -2 - \frac{i}{3} ight)^3} = \frac{x + iy}{27}$Take out the negative sign: $-1 	imes {\left( 2 + \frac{i}{3} ight)^3} = \frac{

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$91$

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(A) Given: ${\left( -2 - \frac{i}{3} ight)^3} = \frac{x + iy}{27}$Take out the negative sign: $-1 	imes {\left( 2 + \frac{i}{3} ight)^3} = \frac{x + iy}{27}$Expand using $(a + b)^3 = a^3 + b^3 + 3a^2b + 3ab^2$:$-1 	imes \left[ 2^3 + \left( \frac{i}{3} ight)^3 + 3(2^2)\left( \frac{i}{3} ight) + 3(2)\left( \frac{i}{3} ight)^2 ight] = \frac{x + iy}{27}$$-1 	imes \left[ 8 - \frac{i}{27} + 4i - \frac{2}{3} ight] = \frac{x + iy}{27}$$-8 + \frac{2}{3} - i\left( 4 - \frac{1}{27} ight) = \frac{x + iy}{27}$Multiply by $27$:$x + iy = 27 	imes \left( -\frac{22}{3} ight) - i 	imes 27 	imes \left( \frac{107}{27} ight)$$x = -198$ and $y = -107$Then $y - x = -107 - (-198) = -107 + 198 = 91$

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

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