### Instruction: The roots of $x^3 + Px^2 + Qx - 19 = 0$ are each one more than the roots of $x^3 - Ax^2 + Bx - C = 0.$  If $A,$ $B,$ $C,$ $P,$ and $Q$ are constants, compute $A + B + C.$.### Response: I notice that the coefficient of $x^3$ in both equations is 1, so they must have the same leading term.
That means I can write $x^3 + Px^2 + Qx - 19 = (x + 1)^3 - Ax^2 + Bx - C$ by expanding the cube and equating coefficients.
Expanding the cube, I get $x^3 + 3x^2 + 3x + 1 - Ax^2 + Bx - C.$
Comparing the coefficients of $x^2,$ I get $3 - A = P,$ so $A = 3 - P.$
Comparing the coefficients of $x,$ I get $3 + B = Q,$ so $B = Q - 3.$
Comparing the constant terms, I get $1 - C = -19,$ so $C = 20.$
Therefore, $A + B + C = (3 - P) + (Q - 3) + 20.$