Tricky Equations - 2

A good solution from Budhaditya Halder in LaTex here:

We note the polynomial $f(x) = (x+a)(x+b)(x+c)-x$ satisfies $f(1) = f(2)=f(3) = 0$ .

$\begin{aligned} \implies f(x) & = (x-1)(x-2)(x-3) \\ \implies f(4) & = 3\times 2 \times 1 = 6 \\ \implies (a+4)(b+4)(c+4) - 4 & = 6 \\ (a+4)(b+4)(c+4) & = \boxed{10} \end{aligned}$

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My Previous Solution

Using the Newton sums method notation for symmetric sums, we have:

$\begin{cases} a + b + c = S_1 \\ ab+bc+ca = S_2 \\ abc = S_3 \end{cases}$

Then,

$\begin{cases} (a+1)(b+1)(c+1)=1 & \implies S_1+S_2+S_3 = 0 & ...(1) \\ (a+2)(b+2)(c+2)=2 & \implies 4S_1+2S_2+S_3 = -6 & ...(2) \\ (a+3)(b+3)(c+3)=3 & \implies 9S_1+3S_2+S_3 = -24 & ...(3) \end{cases}$

\(\begin{array} {} (2)-(1): & 3S_1+S_2 = -6 & ...(4) \\ (3)-(2): & 5S_1+S_2 = -18 & ...(5) \\ (5)-(4): & 2S_1= -12 & \implies S_1 = -6 \\ (4): & -18 + S_2 = -6 & \implies S_2 = 12 \\ (1): & -6 + 12 + S_3 = 0 & \implies S_3 = -6 \end{array} \)

Now, we have:

$\begin{aligned} (a+4)(b+4)(c+4) & = 4^3 + 4^2S_1 + 4S_2 + S_3 \\ & = 64 + 16(-6) + 4(12) - 6 \\ & = 64 - 96 + 48 - 6 \\ & = \boxed{10} \end{aligned}$

We can actually use method of differences here.

It's a polynomial with $P(1) = 1$ $,$ $P(2)=2$ , $P(3)= 3$

Then use the method of finite diffrences to evaluate $P(4)$

Let a+1=X, b+1=Y, c+1=Z. So the equation reduces to:-
(X-1)(Y-1)(Z- 1)=1. $\implies$ XYZ - (XY+YZ+ZX) + (X+Y+Z) = 2.............( * )
XYZ=2......................................................................................................( * * )
(X+1)(Y+1)(Z+1)=3\ \ \ $\implies$ XYZ + (XY+YZ+ZX) + (X+Y+Z) = 2............( * * * )
Adding ( * ) + ( * * * ) , (X+Y+Z)=0, ....and (XY+YZ+ZX)=0. Also XYZ=2.....( * * * * )
(a+4)(b+4)(c+4)=(X+2)(Y+2)(Z+2)= XYZ +2 * (XY+YZ+ZX) +4 * (X+Y+Z) + 8=2+0+0+8= $\large \color{#D61F06}{10}$