Integral part of the product of non-real roots of equation ${\mathrm{x}}^4-4{\mathrm{x}}^3+6{\mathrm{x}}^2-4\mathrm{x}=69$ is _________.

We have ${\mathrm{x}}^4-4{\mathrm{x}}^3+6{\mathrm{x}}^2-4\mathrm{x}=69$

$\begin{array}{l}\Rightarrow {\mathrm{x}}^4-4{\mathrm{x}}^3+6{\mathrm{x}}^2-4\mathrm{x}+1=70\\ \Rightarrow (\mathrm{x}-1{)}^4=70\\ \Rightarrow (\mathrm{x}-1{)}^2=\pm \sqrt{70}\end{array}$

For the product of non-real roots, consider

$\begin{array}{l}(\mathrm{x}-1{)}^2=-\sqrt{70}\\ \Rightarrow {\mathrm{x}}^2-2\mathrm{x}+1+\sqrt{70}=0\end{array}$

Therefore, product is $1+\sqrt{70}$, whose integral part is 9.