Solve the equations:
$a x + b y + c z = k$ ,
$a^{2} x + b^{2} y + c^{2} z = k^{2}$ ,
$a^{3} x + b^{3} y + c^{3} z = k^{3}$ .

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Solution

$Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} a & b & c a^{2} & b^{2} & c^{2} a^{3} & b^{3} & c^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣ = a b c (a - b) (b - c) (c - a)$

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

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

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

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

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Similar questions

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}

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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}

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a x + b y + c z = k

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

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a x + b y + c z = k

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

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