Given that a(y + z) = x, b (z + x) = y , c(x + y) = z, shew that bc + ca+ ab + 2abc = 1.

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Solution

$x - a y - a z = 0..... (i) b x - y + b z = 0.... (i i) c x + c y - z = 0..... (i i i)$

Applying cross multiplicatio on $(i)$ and $(i i)$

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

substituting $x, y$ and $z$ in $(i i i)$

$c k (a + a b) + c (b + a b) - k (1 - a b) = 0 a c + a b c + b c + a b c - 1 + a b = 0 a b + b c + a c + 2 a b c = 1$

Hence proved.

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