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Properties of Conjugate of a Complex Number

Solve the sys...

Question

Solve the system of equations: $z + a y + a^{2} x + a^{3} = 0 z + b y + b^{2} x + b^{3} = 0 z + c y + c^{2} x + c^{3} = 0$

x = a + b + c; y = a b + b c + c a; z = a b c

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x = - (a + b + c); y = a b + b c + c a; z = - a b c

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x = - (a - b + c); y = a b + b c + c a; z = - a b c

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x = a - b + c; y = a b + b c + c a; z = a b c

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Solution

The correct option is B $x = - (a + b + c); y = a b + b c + c a; z = - a b c$
Here, $D = ∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} & a & 1 b^{2} & b & 1 c^{2} & c & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3}$
$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} - b^{2} & a - b & 0 b^{2} - c^{2} & b - c & 0 c^{2} & c & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= (a - b) (b - c) ∣ ∣ ∣ ∣ \begin{matrix} a + b & 1 & 0 b + c & 1 & 0 c^{2} & c & 1 \end{matrix} ∣ ∣ ∣ ∣$
$D = (a - b) (b - c) (a - c)$
Now, $D_{1} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} - a^{3} & a & 1 - b^{3} & b & 1 - c^{3} & c & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3}$
$= - ∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{3} - b^{3} & a - b & 0 b^{3} - c^{3} & b - c & 0 c^{3} & c & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= - (a - b) (b - c) ∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} + a b + b^{2} & 1 & 0 b^{2} + b c + c^{2} & 1 & 0 c^{3} & c & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= - (a - b) (b - c) [a^{2} - c^{2} + (a - c) b]$
$\Rightarrow D_{1} = - (a - b) (b - c) (a - c) (a + b + c)$
Now, $D_{2} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} & - a^{3} & 1 b^{2} & - b^{3} & 1 c^{2} & - c^{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= - ∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} - b^{2} & a^{3} - b^{3} & 0 b^{2} - c^{2} & b^{3} - c^{3} & 0 c^{2} & c^{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= - (a - b) (b - c) ∣ ∣ ∣ ∣ ∣ \begin{matrix} a + b & a^{2} + a b + b^{2} & 0 b + c & b^{2} + b c + c^{2} & 0 c^{2} & c^{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= - (a - b) (b - c) [{a c}^{2} + b c^{2} - a^{2} c - a^{2} b]$
$Δ_{2} = - (a - b) (b - c) (a - c) (a b + b c + a c)$
Now, $D_{3} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} & a & - a^{3} b^{2} & b & - b^{3} c^{2} & c & c^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= - a b c ∣ ∣ ∣ ∣ ∣ \begin{matrix} a & 1 & a^{2} b & 1 & b^{2} c & 1 & c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$

$D_{3} = - a b c ∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} & a & 1 b^{2} & b & 1 c^{2} & c & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$D_{3} = - a b c D$
Hence, $x = \frac{D_{1}}{D} = - (a + b + c)$
$y = \frac{D_{2}}{D} = - (a b + b c + a c)$
$z = \frac{D_{3}}{D} = - a b c$

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