How do i factorize $(z^3-1)(z^2+1)(z+1)+(z^2-1)(z^3+1)(z+1)+(z-1)(z^3+1)(z^2+1)$

Question

How do i find roots of z

$$(z^3-1)(z^2+1)(z+1)+(z^2-1)(z^3+1)(z+1)+(z-1)(z^3+1)(z^2+1)=0$$

Answer to this question is 113(iii)

After expansion i got the same as the expansion in marking scheme but how do i factorize it

Answer 1

Use $z^3-1=(z-1)(z^2+z+1)$ and $z^3+1=(z+1)(z^2-z=1)$. First, add the last two terms together. You get $$(z^3+1)(z-1)(z+1)^2+(z^3+1)(z-1)(z^2+1)=\\(z^3+1)(z-1)(2z^2+2z+2)$$ You can then add the first term: $$(z^3-1)(z^2+1)(z+1)+(z^3+1)(z-1)(2z^2+2z+2)=\\(z-1)(z^2+z+1)(z^2+1)(z+1)+(z+1)(z^2-z+1)(z-1)2(z^2+z+1)=\\(z-1)(z+1)(z^2+z+1)[(z^2+1)+2(z^2-z+1)]=\\(z-1)(z+1)(z^2+z+1)(3z^2-2z+3)=0$$ Solutions are $z=1$, $z=-1$, and the solutions for the two quadratics

Answer 2

If you follow Jyrki Lahtonen's suggestion, you get $$(z^2-1)(3z^4+z^3+4z^2+z+3)=0$$ The thing to notice about the second factor is that the coefficients read backwards and forwards the same way: the polynomial is "palindromic." If we set $$p(z)=3z^4+z^3+4z^2+z+3=0$$ and divide through by $z^4$, we find that whenever $z$ is a root of $p$ then also $\frac1z$ is a root of $p$. Therefore we can say $$(z-a)\left(z-\frac1a\right)(z-b)\left(z-\frac1b\right)=z^4+{z^3\over3}+{4z^2\over3}+{z\over3}+1\tag{1}$$ If we set $$A=a+\frac1a\\B=b+\frac1b,$$ multiply out $(1)$ and compare coefficients, we get $$-A-B=\frac13\\ 2+AB=\frac43$$ We can convert the last two equations into a quadratic and solve for $a$ and $B$, then use the definition of $A$ and $B$ to find $a$ and $b$.

I leave the details to you, but I suggest that you check my calculations before starting. They don't come with any guarantees!