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Question Number 136749 by EDWIN88 last updated on 25/Mar/21

Given system equation

{\begin{cases} x^{2} + 3 xy + y^{2} + 1 = 0 \\ x^{3} + y^{3} - 7 = 0 \end{cases} has solution

(x_{1}, y_{1}) & (x_{2}, y_{2}) for x, y \in R . Find the value

of x_{1}^{2} . y_{2} + x_{2}^{2} . y_{1} .

Answered by MJS_new last updated on 25/Mar/21

x = u - v \land y = u + v

{\begin{cases} 5 u^{2} - v^{2} + 1 = 0 \Rightarrow v^{2} = 5 u^{2} + 1 \\ 2 u^{3} + 6 {u v}^{2} - 7 = 0 \end{cases}

2 u^{3} + 6 u (5 u^{2} + 1) - 7 = 0

u^{3} + \frac{3}{16} u - \frac{7}{32} = 0

u = \frac{1}{2} \lor u = - \frac{1}{4} \pm \frac{\sqrt{6}}{4} i

now {it}^{'} s easy to complete

Answered by bramlexs22 last updated on 26/Mar/21

{\begin{cases} {(x + y)}^{2} + xy + 1 = 0 \\ {(x + y)}^{3} - 3 xy (x + y) - 7 = 0 \end{cases}

set {\begin{cases} x + y = u \\ xy = w \end{cases} \Leftrightarrow {\begin{cases} u^{2} + w + 1 = 0 \\ u^{3} - 3 uw - 7 = 0 \end{cases}

substitute w = - u^{2} - 1

we get u^{3} - 3 u (- u^{2} - 1) - 7 = 0

\Rightarrow {4 u}^{3} + 3 u - 7 = 0

\Rightarrow (u - 1) ({4 u}^{2} + 4 u + 7) = 0

for {4 u}^{2} + 4 u + 7 = 0 rejected

we get u = 1 \land w - 2

so we find {\begin{cases} x + y = 1 \\ xy = - 2 \end{cases}; y = 1 - x

substitute y = 1 - x gives

\Rightarrow x (1 - x) = - 2; x^{2} - x - 2 = 0

(x - 2) (x + 1) = 0 {\begin{cases} x_{1} = - 1 \Rightarrow y_{1} = 2 \\ x_{2} = 2 \Rightarrow y_{2} = - 1 \end{cases}