Observe that $x^{8}-14 x^{4}-8 x^{3}-x^{2}+1 =\left(x^{8}+2 x^{4}+1\right)-\left(16 x^{4}+8 x^{3}+x^{2}\right) =\left(x^{4}+4 x^{2}+x+1\right)\left(x^{4}-4 x^{2}-x+1\right)$. The polynomial $x^{4}+4 x^{2}+x+1=x^{4}+\frac{15}{4} x^{2}+\left(\frac{x}{2}+1\right)^{2}$ has no real roots. On the other hand, let $P(x)=x^{4}-4 x^{2}-x+1$. Observe that $P(-\infty)=+\infty>0, P(-1)=-1<0, P(0)=1>0$, $P(1)=-3<0, P(+\infty)=+\infty>0$, so by the intermediate value theorem, $P(x)=0$ has four distinct real roots, which are precisely the real roots of the original degree 8 equation. By Vieta's formula on $P(x)$, $r_{1}^{2}+r_{2}^{2}+r_{3}^{2}+r_{4}^{2} =\left(r_{1}+r_{2}+r_{3}+r_{4}\right)^{2}-2 \cdot\left(\sum_{i<j} r_{i} r_{j}\right) =0^{2}-2(-4)=8$