FACTORIZATION OF POLYNOMIALS WITH CUBE FORMULA IS ALGEBRAIC IDENTITIES
About "Factorization of Polynomials With Cube Formula in Algebraic Identities"
Factorization of Polynomials With Cube Formula in Algebraic Identities :
Here we are going to see some example problems on factoring polynomials with cube formula in algebraic identities.
Factorization of Polynomials With Cube Formula in Algebraic Identities - Examples
Question 1 :
Factorise the following
(i) 8x3 + 125y3
Solution :
8x3 + 125y3 = (2x)3 + (5y)3
a3 + b3 = (a + b)(a2 - ab + b2)
(2x)3 + (5y)3 = (2x + 5y)((2x)2 - (2x)(5y) + (5y)2)
= (2x + 5y)(4x2 - 10xy + 25y2)
(ii) 27x3 - 8y3
Solution :
27x3 - 8y3 = (3x)3 - (2y)3
a3 - b3 = (a - b)(a2 + ab + b2)
(3x)3 - (2y)3 = (3x - 2y)((3x)2 + (3x)(2y) + (2y)2)
= (3x - 2y)(9x2 + 6xy + 4y2)
(iii) a6 - 64
Solution :
a6 - 64 = (a2)3 - 43
a3 - b3 = (a - b)(a2 + ab + b2)
a = a2 and b = 4
(a2)3 - 43 = (a2 - 4) ((a2)2 + a2 (4) + 42)
= (a2 - 4) ((a4 + 4a2 + 16)
= (a2 - 4) ((a2 + 4)2 - (2a)2)
= (a + 2)(a - 2) ((a2 + 4) + 2a)((a2 + 4) - 2a)
= (a + 2)(a - 2) ((a2 + 2a + 4)((a2 - 2a + 4)
Question 2 :
Factorise the following:
(i) x3 + 8y3 + 6 xy - 1
Solution :
= x3 + 8y3 + 6 xy - 1
= x3 + (2y)3 + (-1)3 - 3 x(2y)(-1)
a3 + b3 + c3 - 3abc = (a + b + c)(a2+b2+c2-ab-bc-ca)
= (x + 2y - 1) (x2 + 4y2 + 1 - 2xy + 2y - x)
(ii) l3 - 8m3 - 27n3 - 18 lmn
Solution :
= l3 - 8m3 - 27n3 - 18 lmn
= l3 + (-2m)3 + (-3n)3 - 3 (1)(-2m)(-3n)
a3 + b3 + c3 - 3abc = (a + b + c)(a2+b2+c2-ab-bc-ca)
= (l - 2m - 3n) (l2 + 4m2 + 9n2 + 2lm - 6mn + 3ln)
After having gone through the stuff given above, we hope that the students would have understood, "Factorization of Polynomials With Cube Formula in Algebraic Identities"
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