Incremental Complex Products

Let $P(x)$ be a monic polynomial with roots $a$ , $b$ and $c$ : $P(x)=(x-a)(x-b)(x-c)$

We know that:

$-P(-1)=(a+1)(b+1)(c+1)=1\\ -P(-2)=(a+2)(b+2)(c+2)=2\\ -P(-3)=(a+3)(b+3)(c+3)=3$

By inspection, we say that $P(x)=(x+1)(x+2)(x+3)+x$ .

So,

$\begin{aligned} -P(-4) & =& (a+4)(b+4)(c+4) \\ & =& -((-4+1)(-4+2)(-4+3)+(-4)) \\ & =& \boxed{10} \\ \end{aligned}$

Let $f(x) = (x+a)(x+b)(x+c) = x^3 +(a+b+c)x^2+(ab+bc+ca)x +abc$ $= x^3 +px^2+qx+r$ where $p = a+b+c, q=ab+bc+ca$ and $r = abc$ . $\\$ Now $f(1) = 1 \Rightarrow p+q+r = 0 ---------(1) \\ f(2) = 2 \Rightarrow 4p+2q+r = -6 -------(2) \\ f(3) = 3 \Rightarrow 9p+3q+4 = -24 -------(3) \\ Solving (1),(2) and (3), p = -6, q = 12 , r = -6. \\ Hence f(x) = x^3 -6x^2+12x-6. \\ Thus f(4) = 10$

Firstly, let's expand all of the equations.

$abc + ab + ac + bc + a + b + c + 1 = 1 \,(I)$

$abc + 2ab + 2ac + 2bc + 4a + 4b + 4c + 8 = 2\,(II)$

$abc + 3ab + 3ac + 3bc + 9a + 9b + 9c + 27 = 3\,(III)$

Subtracting $(I)$ from $(II)$ and $(II)$ from $(III)$ , we get:

$ab + ac + bc + 3a + 3b + 3c + 7 = 1\,(IV)$

$ab + ac + bc + 5a + 5b + 5c + 19 = 1\,(V)$

Subtracting $(IV)$ from $(V)$ :

$2a + 2b + 2c + 12 = 0$

$a + b + c = -6\,(VI)$

Substituting $(VI)$ in $(IV)$ , we get:

$ab + ac + bc + 3*-6 + 7 = 1$

$ab + ac + bc - 18 + 7 = 1$

$ab + ac + bc = 12\,(VII)$

Substituting $(VI)$ and $(VII)$ in $(I)$ :

$abc + 12 - 6 + 1 = 1$

$abc = -6\,(VIII)$

Now we have all of the information we need:

$(a + 4)(b + 4)(c + 4) = abc + 4(ab + ac + bc) + 16(a + b + c) + 64$

$= -6 + 4*12 + 16*-6 + 64 = -6 + 48 - 96 + 64 = 112 - 102 = 10$ .

Let $a+1=x, b+1=y$ , and $c+1=z$ . So $xyz=1$ , $(x+1)(y+1)(z+1)=2$ , $(x+2)(y+2)(z+2)=3$ , and now I want to find $(x+3)(y+3)(z+3)$ .

Expand $(x+1)(y+1)(z+1)=2$ and $(x+2)(y+2)(z+2)=3$ to $xyz+xy+xz+yz+x+y+z+1=2$ , and $xyz+2xy+2xz+2yz+4x+4y+4x+8=3$ .

Then substitute $xyz$ for $1$ and simplify to get $xy+yz+xz+x+y+z=0$ , and $xy+xz+yz+2x+2y+2z=-3$ .

Subtract the first from the second two to get: $x+y+z=-3$ .

Then get $xy+xz+yz=3$ by sustituting $x+y+z=-3$ into the first equation. Expand what we want to find to get: $(x+3)(y+3)(z+3) = xyz+3(xy+xz+yz)+9(x+y+z)+27$ . Substitute out $xyz, xy+yz+xz$ , and $x+y+z. 1+3(3)+9(-3)+27=10$ .

Let $x = a+2$ , $y = b+2$ , and $z = c+2$ . So the system is equivalent to $(x-1)(y-1)(z-1) = 1; xyz = 2; (x+1)(y+1)(z+1) = 3,$ $2-(xy+yz+zx)+x+y+z-1 = 1, xyz = 2; 2+(xy+yz+zx)+(x+y+z)+1 = 3,$ $(x+y+z)-(xy+yz+zx) = 0, xyz = 2; (x+y+z)+(xy+yz+zx) = 0,$ $x+y+z = xy+yz+zx = 0; xyz = 2.$ Now, $(a+4)(b+4)(c+4) = (x+2)(y+2)(z+2) = xyz+2(xy+yz+zx)+4(x+y+z)+8 = 2+2\cdot 0+4\cdot 0+8 = 10.$