Let a is a real number satisfying ${\mathrm{a}}^3+\frac{1}{{\mathrm{a}}^3}=18$. Then the value of ${\mathrm{a}}^4+\frac{1}{{\mathrm{a}}^4}$is ____.

Let $\left(\mathrm{a}+\frac{1}{\mathrm{a}}\right)=\mathrm{t}$
$\Rightarrow {\mathrm{a}}^3+\frac{1}{{\mathrm{a}}^3}=18$
${\mathrm{t}}^3-3\mathrm{t}-18=0 \left(1\right)$
t = 3 satisfies (1)
Hence factorizing (1)
$(\mathrm{t}-3)\left({\mathrm{t}}^2+3\mathrm{t}+6\right)=0$
t = 3 only is the solution
$\therefore \mathrm{a}+\frac{1}{\mathrm{a}}=3\Rightarrow {\mathrm{a}}^2+\frac{1}{{\mathrm{a}}^2}=7$
$\Rightarrow {\mathrm{a}}^4+\frac{1}{{\mathrm{a}}^4}=47$