Algebraic Identities
(a + b)(a - b) = a2 - b2
(a + b)2 = a2 + b2 + 2ac
(a - b)2 = a2 + b2 - 2ac
(a + b)3 = a3 + b3 + 3a2b + 3ab2
(a - b)3 = a3 - b3 - 3a2b + 3ab2
Example 1: Simplify (3u + 5w)(3u – 5w)
Using the algebraic identities (a + b)(a - b) = a2 - b2, we substitute a for 3u and b for 5w.
(3u + 5w)(3u – 5w)
= (3u)2 – (5w)2
= 9u2 – 25w2
Thus (3u + 5w)(3u – 5w) = 9u2 – 25w2
Example 2 : Using the algebraic identities to simplify (3a + 7b)2
Using (a+b)2 = a2+2ab+b2
In this case we need to substitute 3a for a as well as 7b for b
(3a + 7b)2
= (3a)2 + 2(3a)(7b) + (7b)2
= 9a2+ 42ab + 49b2
Thus (3a + 7b)2 = 9a2+ 42ab + 49b2
Example 3: Simplify (5a – 7b)2
Using (a-b)2 = a2-2ab+b2 we have:
(5a – 7b)2
= (5a)2 – 2(5a) (7b) + (7b)2
= 25a2 – 70ab + 49b2.
Thus (5a – 7b)2 = 25a2 – 70ab + 49b2
Example 4 : Expand (2x + 1)3
Using identity (a + b)3 = a3 + b3 + 3a2b + 3ab2 we have:
(2x + 1)3
= (2x)3 + (1)3+ 3(2x)(1)(2x + 1)
= 8x3 + 1 + 6x(2x + 1)
= 8x3 + 12x2 + 6x + 1
Example 5: Expand (2x - 3y)3.
Using identity (a - b)3 = a3 - b3 - 3a2b + 3ab2 we have:
(2x - 3y)3
= (2x)3 - (3y)3 - 3(haha2x)(3y)(2x - 3y)
= 8x3 - 27y3 - 18xy(2x - 3y)
= 8x3 - 27y3 - 36x2y + 54xy2
Example 6: If the values of a+b and ab are 4 and 1 respectively, find the value of a3 + b3.
a3+ b3
= (a+b)3 – 3ab(a+b)
= (4)3 – 3(1)(4)
= 64 – 12
= 52
Example 7: Factorize 27x3 + y3 + z3 - 9xyz
27x3 + y3 + z3 - 9xyz
= (3x)3 + (y)3 + (z)3 - 3(3x)(y)(z)
= (3x + y + z){(3x)2 + (y)2 + (z)2 - (3x)(y) - (y)(z) - (z)(3x)}
Using identity x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - zx) = (3x + y + z)(9x2 + y2 + z2 - 3xy - yz - 3zx)