Solve the following equations :
$z + a y = a^{2} x + a^{3} = 0, z + b y + b^{2} x + b^{3} = 0, z + c y + c^{2} x + c^{3} = 0.$

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Solution

The given equations show that the polynomial

$f (t) = t^{3} + x t^{2} + y t + z$

Vanishes at three different values of $t$ , namely at $t = a, t = b,$ and $t = c$ .

Set up a difference

$t^{3} + x t^{2} + y t + z - (t - a) (t - b) (t - c) .$

This difference also becomes zero at $t = a, b, c .$

Expending this expression in powers of $t,$ we get

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

This second degree equation in $t$ vanishes at three difference value of $t$ and is therefore identically zero, and so each of its coefficient must vanish separately.

Hence

$x + a + b + c = 0, y - a b - b c - c a = 0$

and $z + a b c = 0$

$∴ x = - (a + b + c), y = a b + b c + c a,$ and $z = - a b c .$

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Similar questions

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 y + a^{2} z = a^{3}, x + b y + b^{2} z = b^{3}, x + c y + c^{2} z = c^{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 .

a^{3} + b^{3} + c^{3} - 3 a b c = 0 .

a, b, c

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x + a y + a^{2} z = a^{3}

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x + c y + c^{2} z = c^{3}

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