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?

\begin{align} 
x^8-14x^4-8x^3-x^2+1&=0
\tag{1}\label{1}
.
\end{align}  
\eqref{1} can be factored as
\begin{align}
f_1(x)f_2(x)=0
,\\ 
f_1(x)&=x^4+4x^2+x+1
\tag{2}\label{2}
,\\
f_2(x)&=x^4-4x^2-x+1
\tag{3}\label{3}
.
\end{align}  
\begin{align}
f_1(x)&=x^4+3x^2+(x^2+x+\frac14)-\tfrac14+1
\\
&=x^4+3x^2+\tfrac14(2x+1)^2+\tfrac34
>0\quad \forall x\in\mathbb{R}
.
\end{align}
We can also found that 
factor $f_2(x)$ has four distinct real roots,
for example, observing that
\begin{align}
f_2(-2)&=3>0
,\\
f_2(-1)&=-1<0
,\\
f_2(0)&=1>0
,\\
f_2(1)&=-3<0
,\\
f_2(3)&=43>0
,
\end{align}
Now consider 
\begin{align}
f_2(x)f_2(-x)&=
(x^2)^4-8(x^2)^3+18(x^2)^2-9(x^2)+1
\tag{4}\label{4}
.
\end{align}
Expression \eqref{3}
is a polynomial in $x^2$,
its roots are squares of the roots of\eqref{1},
hence the sought sum 
of the squares of distinct real roots of  \eqref{1}
is $8$ (negated coefficient at $(x^2)^3$ in \eqref{4}).
