Question

if x^2-yz/a=y^2-zx/b=z^2-xy/c , then prove that (a+b+c) (x+y+z) =ax+by+cz​

Answer 1

Answer:

Step-by-step explanation:

Given:

 [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)

hope it helps

have a good day ahead :)