If x 2 -yz a - y 2 -zx b - z 2 -xy c - show that x-y-z a-b-c -ax-by-cz-

x ^ 2 yz a = y ^ 2 zx b = z^ 2 xy c = x ^ 2 yz+ y ^ 2 zx+z^ 2 xy a+b+c = x(x ^ 2 yz)+y( y ^ 2 zx)+z(z^ 2 xy) ax+by+cz = x ^ 3 ...

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Product Law

If x 2 -yz ...

Question

If $\frac{x^{2} - y z}{a} = \frac{y^{2} - z x}{b} = \frac{z^{2} - x y}{c},$ show that $(x + y + z) (a + b + c) = a x + b y + c z .$

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\frac{a x}{y} + \frac{b y}{x} = c, \frac{b y}{z} + \frac{c z}{y} = a, \frac{c z}{x} + \frac{a x}{z} = b,

\frac{y z}{x^{2}} + \frac{z x}{y^{2}} + \frac{x y}{z^{2}}

(a^{2} + b^{2} + c^{2} - b c - c a - a b) (x^{2} + y^{2} + z^{2}

- y z - z x - x y)

= (X^{2} + Y^{2} + Z^{2} - Y Z - Z X - X Y)

x^{2} + y^{2} + z^{2} - x y - y z - z x = \frac{1}{2} [(x - y)^{2} + (y - z)^{2} + (z - x)^{2}]

A = x y, B = y z

C = z x

x^{2} y^{2} z^{2}

\frac{x}{b + c - a} = \frac{y}{c + a - b} = \frac{z}{a + b - c},

\frac{x + y + z}{a + b + c} = \frac{x (y + z) + y (z + x) + z (x + y)}{2 (a x + b y + c z)} .

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