If the roots of the equation $(\lambda+1)x^3-(\lambda^2+2)x^2-(\lambda^2+2)x +(\lambda+1)=0; \lambda \epsilon R$ are in a $A.P$. with non-zero common difference and one of the roots is independent of $\lambda$, then the sum of all possible values of $\lambda$ is $m$. Find the value of $4m$.

The equation is (lambda+1)x^3 - (lambda^2+2)x^2 - (lambda^2+2)x + (lambda+1) = 0. If x= -1, the equation becomes -(lambda+1) - (lambda^2+2) + (lambda^2+2) + (lambda+1) = 0. So x = -1 is a root independent of lambda. Let roots be a/r, a, ar. Product of roots = -d/a = -(lambda+1)/(lambda+1) = -1. So a^3 = -1, a = -1. Since -1 is a root, the other roots are -1/r and -r. Sum = -1 - 1/r - r = (lambda^2+2)/(lambda+1). This leads to a quadratic in lambda.