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Identity: a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
How is this identity obtained? Let's see how. Taking RHS of the identity: (a + b + c)(a2 + b2 + c2 - ab - bc - ca ) Multiply each term of first polynomial with every term of second polynomial, as shown below: = a(a2 + b2 + c2 - ab - bc - ca ) + b(a2 + b2 + c2 - ab - bc - ca ) + c(a2 + b2 + c2 - ab - bc - ca ) = { (a * a2) + (a * b2) + (a * c2) - (a * ab) - (a * bc) - (a * ca) } + {(b * a2) + (b * b2) + (b * c2) - (b * ab) - (b * bc) - (b * ca)} + {(c * a2) + (c * b2) + (c * c2) - (c * ab) - (c * bc) - (c * ca)} Now, solve multiplication in curly braces and we get: = a3 + ab2 + ac2 – a2b - abc - a2c + a2b + b3 + bc2 - ab2 – b2c - abc + a2c + b2c + c3 - abc – bc2 - ac2 Rearrange the terms and we get: = a3 + b3 + c3 + a2b – a2b + ac2- ac2 + ab2 - ab2 + bc2 – bc2 + a2c - a2c + b2c – b2c - abc - abc - abc Above highlighted like terms will be subtracted and we get: = a3 + b3 + c3 - abc - abc - abc Adding like terms i.e (-abc) and we get: = a3 + b3 + c3 – 3abc Hence, a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca) Following are a few applications to this identity. Example: Solve 8a3 + 27b3 + 125c3 – 90abc Solution: This proceeds as: Given polynomial (8a3 + 27b3 + 125c3 – 90abc) can be written as: (2a)3 + (3b)3 + (5c)3 – 3(2a)(3b)(5c) And this represents identity: a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca) Where a = 2a, b = 3b and c = 5c Now apply values of a, b and c on the L.H.S of identity i.e. a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca) and we get: = (2a + 3b + 5c) { (2a)2 +(3b)2 + (5c)2 – (2a)(3b) – (3b)(5c) – (5c)(2a) } Expand the exponential forms and we get: = (2a + 3b + 5c) {4a2 +9b2 + 25c2 – (2a)(3b) – (3b)(5c) – (5c)(2a)} Solve brackets and we get: = (2a + 3b + 5c) {4a2 +9b2 + 25c2 – 6ab – 15bc – 10ca} Hence, 8a3 + 27b3 + 125c3 – 90abc = (2a + 3b + 5c) (4a2 + 9b2 + 25c2 – 6ab – 15bc – 10ca) |