$|\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}{|}^2+|\overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{B}}{|}^2=$

Let $\mathrm{\theta }$  be angle between vectors $\overrightarrow{\mathrm{A}}\mathrm{and}\overrightarrow{\mathrm{B}}$

 $\therefore |\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}|=\mathrm{ABsin\theta }$

and  $|\overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{B}}|=\mathrm{ABcos\theta }$

$|\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}{|}^2=|\overrightarrow{\mathrm{A}.}\overrightarrow{\mathrm{B}}{|}^2={\left(\mathrm{ABsin\theta }\right)}^2+{\left(\mathrm{ABcos\theta }\right)}^2$

= ${\mathrm{A}}^2{\mathrm{B}}^2{\sin }^2\mathrm{\theta }+{\mathrm{A}}^2{\mathrm{B}}^2{\cos }^2\mathrm{\theta } = {\mathrm{A}}^2{\mathrm{B}}^2$