If the roots of the equation x³ + Px² + Ox – 19 = 0 are each one more than the roots of the equation x³ – Ax² + Bx – C = 0, where A, B, C, P & Q are constants then the value of A + B + C =

If the roots of the second polynomial are r, s and t, then substituting x = u - 1 into the first equation gives the roots r+1, s+1 and t+1 for u. The equation (u-1)^3 - A(u-1)^2 + B(u-1) - C = 0 simplifies to u^3 - (A+3)u^2 + (3+2A+B)u - (1+A+B+C) = 0. Matching this to the first equation u^3 + Pu^2 + Qu - 19 = 0, we get the constant term 1+A+B+C = 19. Solving this gives A+B+C = 18.

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