Revision sem1 q3

given,
x²-yza=y²-zxb
→ba=y²-zxx²-yz.....(1)
also,
y²-zxb=z²-xyc
→bc=y²-zxz²-xy.....(2)
therefore,from (1) and (2),we get,
a=x²-yz
b=y²-zx
c=z²-xy
now,solving each side by substituting the value of a,b,c..
R.H.S.
ax+by+cz
=x(x²-yz)+y(y²-yz)+z(z²-xy)
=x³-xyz+y³-xyz+z³-xyz
=x³+y³+z³-3xyz

now,L.H.S.
(a+b+c)(x+y+z)
=(x²-yz+y²-zx+z²-xy)(x+y+z)
=(x+y+z)(x²+y²+z²-xy-yz-zx)
=R.H.S.(by the 7^thidentity).....

hence,proved..

let x2-yz/a=y2-xy/b=z2-xy/c=k
so k(a+b+c)=x2+y2+z2-(xy+yz+zx)
now, x2-yz/z=k
so, x(x2-yz)/ax=k
or kax=x3-xyz
similarly
kby=y3-xyz
kcz=z3-xyz
now: k(ax+by+cz)=x3+y3+z3-3xyz
=(x+y+z)(x2+y2+z2-xy-yz-zx-3xyz+3xyz)
k(ax+by+cz)=(x+y+z)[k(a+b+c)]
k(ax+by+cz)=k(x+y+z)(a+b+c)
hence ax+by+cz+(x+y+z)(a+b+c)
prooved.......