Solve : $(l + m)^{2} - 4 l m$

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Solution

Given, $(l + m^{2}) - 4 l m$

$= l^{2} + m^{2} + 2 l m - 4 l m$

$= l^{2} + m^{2} - 2 l m$ $[∵ (a - b)^{2} = a^{2} + b^{2} - 2 a b]$

$= (l - m)^{2}$

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(l + m)^{2} - 4 l m

(l + m)^{2} - 4 l m

{(l + m)}^{2} - 4 l m

Factorise:
$(l + m)^{2} - 4 l m$

(l + m)^{2} - 4 l m

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