Identity: (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 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) = a^{2} + ab + ac + ab + b^{2} + bc + ac + bc + c^{2} Rearrange the terms and we get: = a^{2} + b^{2} + c^{2} + ab + ab + bc + bc + ac + ac On adding like terms and we get: = a^{2} + b^{2}+ c^{2} + 2ab + 2bc + 2ca Hence, in this way we obtain the identity i.e. (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 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} = a^{2} + b^{2} + c^{2} + 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: = 16p^{2} + 25q^{2} + 4r^{2} + 2(4p)(5q) + 2(5q)(2r) + 2(2r)(4p) Solve brackets and we get: = 16p^{2} + 25q^{2} + 4r^{2} + 40pq + 20qr + 16rp Hence, (4p + 5q + 2r)^{2} = 16p^{2} + 25q^{2} + 4r^{2} + 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} = a^{2} + b^{2} + c^{2} + 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: = 4x^{2} + 16y^{2} + 9z^{2} + 2(2x)(4y) + 2(4y)(3z) + 2(3z)(2x) Solve multiplication process and we get: = 4x^{2} + 16y^{2} + 9z^{2} + 16xy + 24yz + 12zx Hence, (2x + 4y + 3z)^{2} = 4x^{2} + 16y^{2} + 9z^{2} + 16xy + 24yz + 12zx |