# (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

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## (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

 Identity: (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca How is this identity obtained? Let's see how. Taking LHS of the identity: (a + b +c)2 This can also be written as: = (a + b + c) (a + b + c) Multiply as we do multiplication of trinomials and we get: = a(a + b + c) + b(a + b + c) + c(a + b + c) = a2 + ab + ac + ab + b2 + bc + ac + bc + c2 Rearrange the terms and we get: = a2 + b2 + c2 + ab + ab + bc + bc + ac + ac On adding like terms and we get: = a2 + b2+ c2 + 2ab + 2bc + 2ca Hence, in this way we obtain the identity i.e. (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca Following are a few applications to this identity. Example 1: Solve (4p + 5q + 2r)2 Solution: This proceeds as: Given polynomial (4p + 5q + 2r)2 represents identity (a + b + c)2 Where a = 4p, b = 5q and c = 2r Now, apply values of a, b and c on the identity i.e. (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca and we get: (4p + 5q + 2r)2 = (4p)2 + (5q)2 + (2r)2 + 2(4p)(5q) + 2(5q)(2r) + 2(2r)(4p) Expand the exponential forms and we get: = 16p2 + 25q2 + 4r2 + 2(4p)(5q) + 2(5q)(2r) + 2(2r)(4p) Solve brackets and we get: = 16p2 + 25q2 + 4r2 + 40pq + 20qr + 16rp Hence, (4p + 5q + 2r)2 = 16p2 + 25q2 + 4r2 + 40pq + 20qr + 16rp Example 2: Solve (2x + 4y + 3z)2 Solution: This proceeds as: Given polynomial (2x + 4y + 3z)2 represents identity (a + b + c)2 Where a = 2x, b = 4y and c = 3z Now apply values of a, b and c on the identity i.e. (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca and we get: (2x + 4y + 3z)2 = (2x)2 + (4y)2 + (3z)2 + 2(2x)(4y) + 2(4y)(3z) + 2(3z)(2x) Expand the exponential forms and we get: = 4x2 + 16y2 + 9z2 + 2(2x)(4y) + 2(4y)(3z) + 2(3z)(2x) Solve multiplication process and we get: = 4x2 + 16y2 + 9z2 + 16xy + 24yz + 12zx Hence, (2x + 4y + 3z)2 = 4x2 + 16y2 + 9z2 + 16xy + 24yz + 12zx