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

Proved below.

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.

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)