 Solution

For the given symmetric matrix, $(1,1{)}^T$ is an eigenvector with eigenvalue $1$, and $(1,-1{)}^T$ is an eigenvector with eigenvalue $27$. Normalize them to obtainTake the real cube roots of the two diagonal entries and transform back:It is a real symmetric matrix, and $C^3=P\mathrm{diag}(1,27)P^T=A$ because $P^{T}P=I$. Direct multiplication also gives $C^2=\left(\begin{array}{cc}5& -4\\ -4& 5\end{array}\right)$ and then $C^3=\left(\begin{array}{cc}14& -13\\ -13& 14\end{array}\right)$.