Find x ?_?

This may be a bit confusing. So the problem said:

$\frac{x+x^{2}+ x^{3}+x^{4}+x^{5}}{3} = 121$

$x+x^{2}+x^{3}+x^{4}+x^{5} = 363$

$x+x^{2}+x^{3}+x^{4}+x^{5} - 363 = 0$

Now we will "cheat" a bit:

$x + x^{2}+ x^{3}+x^{4}+x^{5}-27-12-324= 0$

$(x^{3} - 27) +(x^{2}+x-12) + (x^{4}+x^{5}-324) =0$

$(x^{3}-3^{3})+(x^{2}+4x-3x-12)+(x^{4}+x^{5}-324) =0$

$(x-3)(x^{2}+3x+9)+(x-3)(x+4)+(x-3)(x^{4}+4x^{3}+12x^{2}+36x+108) = 0$

Now we can see $x-3$ on each term, we should have:

$(x-3)(x^{2}+3x+9+x+4+x^{4}+4x^{3}+12x^{2}+36x+108) =0$

$(x-3)(x^{4}+4x^{3}+13x^{2}+40x+121)=0$

⇒ $x-3 =0$ or $(x^{4}+4x^{3}+13x^{2}+40x+121)=0$

$x-3=0$ ⇒ $x=3$

$(x^{4}+4x^{3}+13x^{2}+40x+121)=0$ ⇒ $x =$ I don't know.

So $x=3$