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Identity: (a + b)3 = a3 + b3 + 3ab(a + b)
How is this identity obtained? Let's see how. Taking LHS of the identity: (a + b)3 This can also be written as: = (a + b) (a + b) (a + b) Now, multiply first two binomials as shown below: = { a(a + b) + b(a + b) } (a + b) = { a2 + ab + ab + b2 } (a + b) On rearranging the terms in curly braces we get: = { a2 + b2 + ab + ab } (a + b) = { a2 + b2 + 2ab } (a + b) Now, multiply trinomial with binomial as shown below: = a2(a + b) + b2(a + b) + 2ab(a + b) = a3 + a2b + ab2 + b3 + 2ab(a + b) = a3 + b3 + a2b + ab2 + 2ab(a + b) = a3 + b3 + a2b + ab2 + 2ab(a + b) Take ab common from the above red highlighted terms and we get: = a3 + b3 + ab(a + b) + 2ab(a + b) On adding like terms and we get: = a3 + b3 + 3ab(a + b) On solving it further we get: a3 + b3 + 3a2b + 3ab2 Hence, in this way we obtain the identity i.e. (a + b)3 = a3 + b3 + 3ab(a + b) = a3 + b3 + 3a2b + 3ab2 Example 1: Solve (3a + 2b)3 Solution: This proceeds as: Given polynomial (3a + 2b)3 represents the identity (a + b)3 Where a = 3a and b = 2b Now substitute values of a and b in the identity i.e. (a + b)3 = a3 + b3 + 3ab(a + b) and we get: (3a + 2b)3 = (3a)3 + (2b)3 + 3(3a) (2b)(3a + 2b) Expand the exponential forms and we get: = 27a3 + 8b3 + 3(3a) (2b)(3a + 2b) Solve multiplication process and we get: = 27a3 + 8b3 + 18ab(3a + 2b) Hence, (3a + 2b)3 = 27a3 + 8b3 + 18ab(3a + 2b) Example 2: Solve (5x + 4y)3 Solution: This proceeds as: Given polynomial (5x + 4y)3 represents identity i.e. (a + b)3 Where a = 5x and b = 4y Now apply values of a and b on the identity i.e. (a + b)3 = a3 + b3 + 3a2b + 3ab2 and we get: (5x + 4y)3 = (5x)3 + (4y)3 + 3(5x)2 (4y) + 3(5x) (4y)2 Expand the exponential forms and we get: = 125x3 + 64y3 + 3(25x2)(4y) + 3(5x) (16y2) Solve multiplication process and we get: = 125x3 + 64y3 + 300x2y + 240xy2 Hence, (5x + 4y)3 = 125x3 + 64y3 + 300x2y + 240xy2 |
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(a + b)2 |
(a - b)2 |
a2 - b2 |
(a - b)3 |
(a + b + c)2 |
a3 + b3 + c3 - 3abc |
Linear Equations in one variable |