Factorize the Complex

If $x^3-1$ is a factor of $f(x) = x^6+ax^4+bx^3+cx^2+3x+2$ . Then $1$ , $\omega$ and $\omega^2$ , where $\omega$ is the third root of unity, are the roots of $f(x)$ . Then, we have:

$\begin{cases} x=1, & \implies 1 + a + b + c + 3 + 2 = 0 & ...(1) \\ x=\omega, & \implies 1 + a\omega + b + c\omega^2 + 3\omega + 2 = 0 & ...(2) & \small \color{#3D99F6} \text{Note that: } \omega^3 = 1 \\ x=\omega^2, & \implies 1 + a\omega^2 + b + c\omega + 3\omega^2 + 2 = 0 & ...(3) \end{cases}$

Since $\omega^2+\omega+1 = 0$ , $\implies (1)+(2)+(3): 3 + (0)a + 3b + (0)c + 3(0) + 6 = 0$ $\implies b = -3$ . Then then system of equations becomes:

$\begin{cases} a + c + 3 = 0 & ...(1a) \\ a\omega + c\omega^2 + 3\omega = 0 & ...(2a) \\ a\omega^2 + c\omega + 3\omega^2 = 0 & ...(3a) \end{cases}$

Then $(2a) \times \omega$ and $(3a) \times \omega^2$ $\implies \begin{cases} a + c + 3 = 0 & ...(1a) \\ a\omega^2 + c + 3\omega^2 = 0 & ...(2b) \\ a\omega + c + 3\omega = 0 & ...(3b) \end{cases}$

$(1a)+(2b)+(3b): (0)a + 3c + 3(0) = 0 \implies c = 0$ . From $(1a): \implies a = -3$ .

Therefore, $ab+bc+ca = (-3)(-3) + 0+0 = \boxed 9$ .

We can see that $x^6+ax^4+bx^3+cx^2+3x+2=(x^3-1)(x^3-3x-2)=x^6-3x^4-3x^3+3x+2$ . Now $a=b=-3$ and $c=0$ , and the answer is $\boxed{9}$

Maybe there's an easier solution, but in this case, the division gives:

$\large \displaystyle x^6 + ax^4 + bx^3 + cx^2 + 3x + 2 = [x^3 + ax + (b-1) ] \cdot [x^3 - 1] + [cx^2 + (3+a)x + (3+b)]$

The remainder $cx^2 + (3+a)x + (3+b)$ should be equal to the null polynomial. That is:

$\color{#3D99F6} \large \displaystyle c = 0, a = - 3, b = - 3, \boxed{\large \displaystyle ab+bc+ac = 9}$

If $x^3-1$ is a factor than we can write $(x^3-1)(\alpha x^3 +\beta x^2+\gamma x+\delta)=x^6+ ax^4+bx^3+cx^2+ 3x+2$ . After the multiplication we get

$\alpha x^6+bx^5+\gamma x^4+(\delta-\alpha)x^3 -\beta x^2-\gamma x-\delta=x^6+ ax^4+bx^3+cx^2+ 3x+2$

We identify the coefficients of the various powers of $x$ :

$\alpha=1$

$\beta = 0$

$\gamma=a$

$\delta-\alpha = b$

$-\beta=c$

$\gamma=-3$

$\delta=-2$

and when we eliminate/substitute variables we get $a=-3$ , $b=-3$ and $c=0$ , so $ab+ac+bc=9$ .

We are in effect told that, for some as yet unknown p,q and r;

$(x^3 -1)(x^3+px^2+qx+r)=x^6+ax^4+bx^3+cx^2+3x+2$

Which we view not as an equation but as an identity. Starting from the fifth and equating the coefficients of descending powers of x yields

$p=0\\q=a\\r-1=b\\-p=c\\-q=3\\-r=2$

These solve easily to give $a=-3\\b=-3\\c=0$ ,

and so

$ab+ac+bc=-3\times-3=\boxed{9}$

$(x^3-1)$ is a factor $\implies (x-1)$ is a factor $\implies 1$ is a root. $\implies a+b+c+6=0$

Also, Division algorithm yields $x^6+ax^4+bx^3+cx^2+3x+2=(x^3-1)\cdot (x^3+ax+(b+1))+(cx^2+(a+3)x+(b+3)) \implies c=0, a=-3, b=-3$ , since the remainder $(cx^2+(a+3)x+(b+3))$ should be $0$ .

$\implies ab+bc+ac=9$

I don't agree that Chew-Seong's solution is the easiest. A simple division shows that the polynomial can be written as $\left[x^3 + ax + (b+1)\right]\left[x^3 - 1\right] + \text{remainder}.$ This remainder must be zero, so the polynomial is exactly equal to $\left[x^3 + ax + (b+1)\right]\left[x^3 - 1\right] = x^6 + ax^4 + bx^3 - ax - (b+1).$ So we see that $x^6 + ax^4 + bx^3 + cx^2 + 3x + 2 = x^6 + ax^4 + bx^3 - ax - (b+1).$ Immediately we conclude $c = 0$ , $a = -3$ , and $b+1 = -2$ (so that $b = -3$ ). Therefore $ab +bc + ac = ab = (-3)(-3) = \boxed{9}.$

(x^3+px^2+qx+r)(x^3-1)=x^6+px^5+qx^4+(r-1)x^3-px^2-qx-r

p=0; q=a; r=b+1; p=c=0; q=-3; r=-2

thus: a=-3; b=-3; c=0