If a + b + c = 0 and $a^2+b^2+c^2=1$, find the value of $\left.(a^4+b^4+c^4\right)$.

Let a, band c be the roots of $x^3+px+q=0$ Here, $s_1=(a+b+c)=0$

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{s}_2=a^2+b^2+c^2=(a+b+c{)}^2-2(ab+bc+ca)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1=0-2(ab+bc+ca)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(ab+bc+ca)=-1/2\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}p=-1/2\end{array}$

Now, 

$\begin{array}{l}{x}^3+px+q=0\\ \Rightarrow x^3=-px-q\\ \Rightarrow x^4=-p{x}^2-qx\\ \Rightarrow \mathrm{\Sigma }{a}^4=-p\mathrm{\Sigma }{a}^2-q\mathrm{\Sigma }a\\ \Rightarrow \mathrm{\Sigma }{a}^4=-p\left(1\right)-q\left(0\right)\\ \Rightarrow \sum a^4=-p\left(1\right)=-\left(-\frac{1}{2}\right)=\frac{1}{2}\\ \Rightarrow \left(a^4+b^4+c^4\right)=\frac{1}{2}\end{array}$

Hence the value of $\left.(a^4+b^4+c^4\right)$ is $\frac{1}{2}$.