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

Open in App

Solution

$a x + b y + c z = 0 a^{2} x + b^{2} y + c^{2} z = 0 \frac{x}{b c^{2} - b^{2} c} = \frac{- y}{a c^{2} - a^{2} c} = \frac{z}{a b^{2} {- a}^{2} b} \frac{x}{b c (b - c)} = \frac{y}{a c (c - a)} = \frac{z}{a b (a - b)} = k \Rightarrow x = k b c (b - c), y = k a c (c - a), z = k a b (a - b) x + y + z + (b - c) (c - a) (a - b) = 0$

Substituting $x, y$ and $z$

$k b c (b - c) + k a c (c - a) + k a b (a - b) + (b - c) (c - a) (a - b) = 0 k {b c^{2} - b^{2} c - a c^{2} + a^{2} c + a b^{2} {- a}^{2} b} + {(b c - a b - c^{2} + a c) (a - b)} = 0 k {b c^{2} - b^{2} c - a c^{2} + a^{2} c + a b^{2} {- a}^{2} b} + {a b c - a^{2} b - a c^{2} + a^{2} c - b^{2} c + a b^{2} - b c^{2} - a b c} = 0 k {b c^{2} - b^{2} c - a c^{2} + a^{2} c + a b^{2} {- a}^{2} b} - {b c^{2} - b^{2} c - a c^{2} + a^{2} c + a b^{2} {- a}^{2} b} = 0 \Rightarrow k - 1 = 0 \Rightarrow k = 1 \Rightarrow x = b c (b - c), y = a c (c - a), z = a b (a - b)$

Suggest Corrections

Similar questions

x + y + z + u = 0

a x + b y + c z + d u = 0

a^{2} x + b^{2} y + c^{2} z + d^{2} u = 0

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

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 = 0 . . . (1),

x + y + z = 0 . . . (2),

b c x + c a y + a b z = (b - c) (c - a) (a - b) . . . (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}

x + y + z = 1

a x + b y + c z = k

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

Join BYJU'S Learning Program