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:

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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)

(Proved)

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