請問有妙解嗎?

\(x^2-7x+1=0\)的解為\(b\)，求\(\displaystyle b^2+\frac{1}{b^2}=\)？

\(\begin{align}
  & {{b}^{2}}+1=7b \\
& b+\frac{1}{b}=7 \\
& {{b}^{2}}+\frac{1}{{{b}^{2}}}={{\left( b+\frac{1}{b} \right)}^{2}}-2=47 \\
\end{align}\)

感謝妙解