= Solution
{c}
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 obtain
$$
P=\frac1{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix},\qquad
A=P\begin{pmatrix}1&0\\0&27\end{pmatrix}P^T.
$$
Take the real cube roots of the two diagonal entries and transform back:
$$
\boxed{C=P\begin{pmatrix}1&0\\0&3\end{pmatrix}P^T
=\begin{pmatrix}2&-1\\-1&2\end{pmatrix}.}
$$
It is a real <symmetric matrix>, and $C^3=P\operatorname{diag}(1,27)P^T=A$ because $P^TP=I$. Direct multiplication also gives $C^2=\begin{pmatrix}5&-4\\-4&5\end{pmatrix}$ and then $C^3=\begin{pmatrix}14&-13\\-13&14\end{pmatrix}$.
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