(x+yi)^3 = ((10(2y+8i)^2 )/(3−i)) find x & y


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Question Number 104657 by bobhans last updated on 23/Jul/20

${(x + y i)}^{3} = \frac{10 {(2 y + 8 i)}^{2}}{3 - i}$ $f i n d x & y$
	Answered by bramlex last updated on 23/Jul/20
	$(x+yi)^3 = ((10(2y+8i)^2 (3+i))/(10)) (x+yi)^3 = 4(3+i)(y+4i)^2 x^3 +3x^2 yi−3xy^2 −y^3 i = 4(3+i)(y^2 +8yi−16) x^3 +3x^2 yi−3xy^2 −y^3 i = 4(3y^2 +24yi−48+ y^2 i−8y−16i) x^3 +3x^2 yi−3xy^2 −y^3 i = 12y^2 +96yi−192+ 4y^2 i−32y−64i) → { ((x^3 −3xy^2 = 12y^2 −192−32y)),((3x^2 yi −y^3 i = 4y^2 i−64i )) :} we get { ((x^3 −3xy^2 = 12y^2 −32y−192)),((3x^2 y−y^3 = 4y^2 −64)) :}$
	${(x + y i)}^{3} = \frac{10 {(2 y + 8 i)}^{2} (3 + i)}{10}$ ${(x + y i)}^{3} = 4 (3 + i) {(y + 4 i)}^{2}$ $x^{3} + 3 x^{2} y i - 3 {x y}^{2} - y^{3} i =$ $4 (3 + i) (y^{2} + 8 y i - 16)$ $x^{3} + 3 x^{2} y i - 3 {x y}^{2} - y^{3} i = 4 (3 y^{2} + 24 y i - 48 +$ $y^{2} i - 8 y - 16 i)$ $x^{3} + 3 x^{2} y i - 3 {x y}^{2} - y^{3} i = 12 y^{2} + 96 y i - 192 +$ $4 y^{2} i - 32 y - 64 i)$ $\to {\begin{cases} x^{3} - 3 {x y}^{2} = 12 y^{2} - 192 - 32 y \\ 3 x^{2} y i - y^{3} i = 4 y^{2} i - 64 i \end{cases}$ $w e g e t$ ${\begin{cases} x^{3} - 3 {x y}^{2} = 12 y^{2} - 32 y - 192 \\ 3 x^{2} y - y^{3} = 4 y^{2} - 64 \end{cases}$
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