Finding the sum of squares of the real roots 
Let $r_1,r_2,r_3,\cdots,r_n$ be the distinct real zeroes of the equation $$x^8-14x^4-8x^3-x^2+1=0.$$ Then $r_1^2+r_2^2+r_3^2+\cdots+r_n^2$ is $$(A)\,3\quad(B)\,14\quad(C)\,8\quad(D)\,16$$

I can get the sum of the squares of all roots using Vieta’s formulae, but I don't know actually how to proceed in this question. 
Do I need to draw a graph and then find the answers or is there a sum trick here which I am not able to see through?

Hint:   the real roots are the roots of the first factor:
$$
\begin{align}
x^8\color{red}{+2x^4-2x^4}-14x^4-8x^3-x^2+1 &= (x^4+1)^2 - x^2(4x+1)^2 \\
 &= (x^4 - 4 x^2 - x+1)(x^4+4x^2+x+1)
\end{align}
$$
