Q2/127) If a, b, c are different and the equations ax+a^2y+(a^3+1)=0 bx+b^2y+(b^3+1)=0 cx+c^2y+(c^3+1)=0 are consistent, prove that abc+1=0...

Let us have a function f(t) = xt+yt^2+(t^3+1)

We have f(a) = 0 , f(b) = 0, f(c) = 0

And f(t) = t^3 + yt^2 + xt + 1 …1

So a,b, c are zeroes of f(t) which is a cubic polynomial

So f(t) = m(t-a)(t-b)(t-c) …2

Comparing coefficient of t^3 from (1) and (2) m = 1 

Now constant term = -mabc = 1 or –abc = 1 or abc + 1 = 0

Proved