If x2 - yza = y2 - zxb = z2 - xyc , then prove that:
(a + b + c)(x + y + z) = ax + by + cz
269given,
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 7thidentity).....
hence,proved..
Astha Gupta it is proved since...RHS=LHS...
Sayantan Hazra From 1 and 2, you cannot directly substitute a = x^2 - yz......you can only say x^2 - yz = ak.......rest is fine :)Astha Gupta ok..i get it......Sayantan Hazra so..no pink here :(Astha Gupta 😱 ðŸ˜....101let 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.......
Sayantan Hazra Good work!! :)