Polynomial Identities

In this page we are going to learn about polynomial identities concept. Algebra is the branch of mathematics concerning the study of the rules of operations and relations, and the constructions and concepts arising from them, including terms, polynomials, equations and algebraic structures. A polynomial is an expression of finite length constructed from variables and constants, using only the operations of addition, subtraction, multiplication, and non-negative, whole-number exponents. In this article we shall discuss polynomial identities.(Source: wikipedia)

Below are the polynomial identities -

(i) Identity for (a + b)3

(a + b)3 = (a + b)(a + b)(a + b)

= a3 + (b + b + b)a2 + (b × b + b × b + b × b)a + b × b × b = a3 + 3a2b + 3ab2 + b3

( a + b)3 ≡ a3 + 3a2b + 3ab2 + b3

Reversing the sides,

a3 + b3 ≡ (a + b)3−3ab(a + b)

Based on the identities for (a+b)3 and (a–b)3, we deduce the following identities:

(a) (a+b)3–3ab(a+b) = a3 + 3a2b + 3ab2 + b3 – 3a2b – 3ab2 = a3+b3.

(a + b)3 − 3ab(a + b)≡a3+ b3

a3 + b3 ≡ (a + b)3−3ab(a + b)

(ii) Identity for (a – b)3

(a – b)3 = [a + (–b)]3

= a3 + 3a2(–b) + 3a(–b)2 + (–b)3= a3 – 3a2b + 3ab2 – b3

(a − b)3 ≡a3−3a2b+ 3ab2−b3


(b) (a – b)3+3ab(a – b) = a3 – 3a2b + 3ab2 – b3 + 3a2b – 3ab2 = a3–b3.

(a − b)3 + 3ab(a − b) ≡ a3 − b3

Reversing the sides,

a3 − b3 ≡ (a −b)3+ 3ab(a−b)

(iii) a3 + b3 = (a + b)3 – 3ab(a + b) = (a + b)(a + b)2 – 3ab(a + b)

= (a + b)[(a + b)2 – 3ab] = (a + b)[(a2 + 2ab + b2) – 3ab]

= (a + b)(a2 – ab + b2)

a3 + b3 ≡ (a + b)(a2 – ab + b2)

Reversing the sides,

(a + b)(a2 – ab + b2) ≡ a3 + b3

(iv) a3 – b3 = (a – b)3 + 3ab(a – b) = (a – b)(a – b)2 + 3ab(a – b)

= (a – b)[(a – b)2 + 3ab] = (a – b)[(a2 – 2ab + b2) + 3ab]

= (a – b)(a2 + ab + b2)

a3 − b3 ≡ (a−b)(a2 + ab + b2)

Reversing the sides,

(a − b)(a2 + ab + b2) ≡ a3 − b3

Examples on polynomial identities

Below are the example problems on polynomial identities -

Example 1:

Expand the (3x+2y)3

Solution:

(3x+2y)3 = (3x)3 + 3(3x)2(2y) + 3(3x)(2y)2 + (2y)3

= 27x3 + 3(9x2)(2y) + 3(3x)(4y2) + 8y3

= 27x3 + 54x2y + 36xy2 + 8y3.


Example 2:

Expand the (2x2–3y)3

Solution:

(2x2– 3y)3 = (2x2)3 – 3(2x2)2 (3y) + 3(2x2) (3y)2 – (3y)3

= 8x6 – 3(4x4)(3y) + 3(2x2)(9y2) – 27y3

= 8x6– 36x4y + 54x2y2 – 27y3