Find the conditions for the roots $\alpha, \beta, \gamma$ of the equation $x^3-ax^2+bx-c=0$ to be in: $(i)$A.P.; $(ii)$G.P.
If the roots are not in A.P. and if $\alpha+\lambda,\ \beta+\lambda,\ \gamma+\lambda$ are in G.P., prove that $\lambda$ is given by a cubic equation.
I've got the solution for $(i)\ 2a^3-9ab+27c=0$ and $(ii)\ a^3c=b^3$
But how to prove the cubic equation for $\lambda$?
Let $p(x)=x^3-ax^2+bx-c$, then the polynomial with roots $\alpha+\lambda, \beta+\lambda, \gamma+\lambda$ is: $$p(x-\lambda) =(x-\lambda)^3-a(x-\lambda)^2+b(x-\lambda)-c$$ $$=x^3-(3\lambda+a)x^2+(3\lambda^2+2a\lambda+b)x-(\lambda^3+a\lambda^2+b\lambda+c)$$
You already know from $(ii)$, the condition on coefficients for these roots to be in GP, so use that and simplify to get the cubic for $\lambda$. i.e. $$(3\lambda+a)^3(\lambda^3+a\lambda^2+b\lambda+c)=(3\lambda^2+2a\lambda+b)^3$$ $$\iff (2a^3+27c-9ab)\lambda^3+(a^4+27ac-3a^2b-9b^2)\lambda^2+(a^3b+9a^2c-6ab^2)\lambda+(a^3c-b^3)=0$$
This remains a cubic as long as $2a^3+27c\neq 9ab$, which as per the condition in $(i)$, is ensured if the original roots are not in A.P.