Evaluate (a+2b+3c)2
a2+4b2+9c2+4ab+12bc+6ac
a2+4b2+9c2+4ab+12bc+12ac
a2+4b2+9c2+8ab+12bc+6ac
a2+4b2+9c3+4ab+12bc+6ac
We use the identity (a+b+c)2 = a2+b2+c2+2ab+2bc+2ca
Replacing the value of a,b,and c with a,2b, and 3c respectively.
We get (a+2b+3c)2 = a2+(2b)2+(3c)2+4ab+12bc+6ac
= a2+4b2+9c2+4ab+12bc+6ac
Expand (4a–2b–3c)2