Question

If (x²-yz)/a = (y²-zx)/b = (z²-xy)/c, then prove that (a+b+c) (x+y+z) = ax+by+cz

Answer 1

Answer:

Step-by-step explanation:

Given:

\frac{(x^2-yz)}{a} = \frac{(y^2-zx)}{b} = \frac{(z^2-xy)}{c} = k [say] [1]

Multiply both numerator and denominators by x, y and z in respective

equations and sum up the numerators/ denominators.

x(x^2-yz) +y(y^2-zx) + z(z^2- xy)= k(ax+by+cz)

x^3 -xyz + y^3-xyz+z^3-xyz= k(ax+by+cz)

x^3+y^3+z^3- 3xyz = k(ax+by+cz)

(x+y+z) (x^2 +y^2+z^2-xy-yz-zx) =k (ax+by+cz)(x+y+z) (x^2-yz +y^2-zx+z^2-xy ) =k (ax+by+cz)

x+y+z) (ka +kb+kc ) =k (ax+by+cz) [ from 1 ]

(x+y+z) (a +b+c ) = (ax+by+cz)

Hence proved

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Answer 2

Step-by-step explanation:

x^2-yz)/a=(y^2-zx)/b=(z^2-xy)/c =k say

Multiply both numerator and denominators by x, y and z in respective equations and sum up the numerators/ denominators.

x(x^2-yz) +y(y^2-zx) + z(z^2- xy)= k(ax+by+cz)

x^3 -xyz + y^3-xyz+z^3-xyz= k(ax+by+cz)

x^3+y^3+z^3- 3xyz= k(ax+by+cz)

(x+y+z) (x^2 +y^2+z^2-xy-yz-zx) =k(ax+by+cz)

(x+y+z) (x^2-yz +y^2-zx+z^2-xy )=k(ax+by+cz)

(x+y+z) (ka +kb+kc )=k(ax+by+cz)

(x+y+z) (a +b+c )=(ax+by+cz)

hence Proved