Upon factorisation using algebraic identities, ${(\mathrm{l}+\mathrm{m})}^2-4\mathrm{lm}$  gives the final result as ________.

We will first expand ${(\mathrm{l}+\mathrm{m})}^2$ using the identity: ${(\mathrm{a}+\mathrm{b})}^2={\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab}$

${(\mathrm{l}+\mathrm{m})}^2-4\mathrm{lm}={\left(\mathrm{l}\right)}^2+{\left(\mathrm{m}\right)}^2+2\times \mathrm{l}\times \mathrm{m}-4\mathrm{lm}$

                           $={\mathrm{l}}^2+{\mathrm{m}}^2+2\mathrm{lm}-4\mathrm{lm}$

                           $={\mathrm{l}}^2+{\mathrm{m}}^2-2\mathrm{lm}$

                           ${=(\mathrm{l}-\mathrm{m})}^2$   [ Using identity: ${(\mathrm{a}-\mathrm{b})}^2={\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab}$ ]