Given: (2l+m)2−8lm
On expanding (2l+m)2using the identity (a+b)2=a2+2ab+b2,we get,
(2l+m)2=(2l)2+4lm+m2
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So, (2l+m)2−8lm=(2l)2+4lm+m2−8lm=(2l)2+(−4lm)+m2
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By comparing this with the sameidentity (a+b)2=a2+2ab+b2,a=2l,b=−m(2l+m)2−8lm=(2l−m)2
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