Complex Numbers

lets think of it as a cubic polynomial $G(x)=(x+a)(x+b)(x+c)$ now, let $F(x)=G(x)-x=(x+a)(x+b)(x+c)-x$ we see that $F(1)=F(2)=F(3)=0$ since 1,2,3 are roots of F(x), it can be written as $F(x)=(x-1)(x-2)(x-3)$ lets compute $F(4)=(4-1)(4-2)(4-3)=3\times 2\times 1=6$ lets solve for G(4) $G(4)-4=F(4)\longrightarrow G(4)=6+4$ $G(4)=10$ and hence $(a+4)(b+4)(c+4)=\boxed{10}$

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.