The equation ${\left({\mathrm{x}}^2+\mathrm{x}+1\right)}^2+1=\left({\mathrm{x}}^2+\mathrm{x}+1\right)\left({\mathrm{x}}^2-\mathrm{x}-5\right)$ for $\mathrm{x}\in (-2,3)$ will have number of solutions, 

${\left({\mathrm{x}}^2+\mathrm{x}+1\right)}^2+1=\left({\mathrm{x}}^2+\mathrm{x}+1\right)\left({\mathrm{x}}^2-\mathrm{x}-5\right)$

Since, ${\mathrm{x}}^2+\mathrm{x}+1>0\mathrm{\forall }\mathrm{x}\in \mathrm{R}$

Dividing both sides by ${\mathrm{x}}^2+\mathrm{x}+1$, we get

$\left({\mathrm{x}}^2+\mathrm{x}+1\right)+\frac{1}{\left({\mathrm{x}}^2+\mathrm{x}+1\right)}={\mathrm{x}}^2-\mathrm{x}-5$

$\therefore \text{ L.H.S. }\ge 2$

But ${\mathrm{x}}^2-\mathrm{x}-5<1\text{ for }\mathrm{x}\in (-2,3)$

Therefore, the equation has no solution.