That's pretty complex

$x^2+x+1=\frac{x^3-1}{x-1}=0$ $\implies x^3+1=0$ $x^3=1$ Leave the above equation as $Equation 1$ . Next, we have: $x^2+x+1=0$ $x^2+x=-1$ Leave the equation as $Equation 2$ . Given the question is: $x^{49}+x^{50}+x^{51}+x^{52}+x^{53}$ $=x^{49}\left(1+x+x^2\right)+x^{51}\left(x+x^2\right)$ Substitute $Equation 2$ and the given equation in the above equation and we got: $x^{51}\left(-1\right)$ Since $x^3=1$ and $\text{51 is a multiple of 3}$ , we can say that $x^{51}=1$ . Thus, $1\left(-1\right)$ $=\color{#3D99F6}\text{-1}$

$x^{49} + x^{50} + x^{51} + x^{52} + x^{53} = x^{49}(1 + x + x^{2}) + x^{51}(1 + x + x^{2}) - x^{51} = -x^{51} = -(x^{3})^{17}$ since $1 + x + x^{2} = 0$ .

Next, note that $x^{2} + x + 1 = 0 \Longrightarrow x^{3} + x^{2} + x = 0 \Longrightarrow x^{3} = -(x^{2} + x) = -(-1) = 1$ .

So finally the desired answer is $-(x^{3})^{17} = -(1)^{17} = \boxed{-1}$ .