Solve the equations:
$x + y + z = 1$ ,
$a x + b y + c z = k$ ,
$a^{2} x + b^{2} y + c^{2} z = k^{2}$ .

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Solution

$Δ = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a & b & c a^{2} & b^{2} & c^{2} \end{matrix} ∣ ∣ ∣ ∣ = (a - b) (b - c) (c - a)$

$Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 k & b & c k^{2} & b^{2} & c^{2} \end{matrix} ∣ ∣ ∣ ∣ = - (b - k) (c - k) (b - c)$

$Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a & k & c a^{2} & k^{2} & c^{2} \end{matrix} ∣ ∣ ∣ ∣ = - (a - k) (c - k) (c - a)$

$Δ_{3} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a & b & k a^{2} & b^{2} & k^{2} \end{matrix} ∣ ∣ ∣ ∣ = - (a - k) (b - k) (a - b)$

$x = \frac{Δ_{1}}{Δ} = - \frac{(b - k) (c - k)}{(a - b) (c - a)}; y = \frac{Δ_{2}}{Δ} = - \frac{(a - k) (c - k)}{(a - b) (b - c)}; z = \frac{Δ_{3}}{Δ} = - \frac{(a - k) (b - k)}{(b - c) (c - a)}$

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Similar questions

x + y + z + u = 1

a x + b y + c z + d u = k

a^{2} x + b^{2} y + c^{2} z + d^{2} u = k^{2}

a^{3} x + b^{3} y + c^{3} z + d^{3} u = k^{3}

x + y + z = 1

a x + b y + c z = k

a^{2} x + b^{2} y + c^{2} z = k^{2}

a x + b y + c z = k

a^{2} x + b^{2} y + c^{2} z = k^{2}

a^{3} x + b^{3} y + c^{3} z = k^{3}

a x + b y + c z = k

a^{2} x + b^{2} y + c^{2} z = k^{2}

a^{3} x + b^{3} y + c^{3} z = k^{3}

a x + b y + c z = a^{2} x + b^{2} y + c^{z} = 0, x + y + z + (b - c) (c - a) (a - b) = 0 .

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