Factorization of Cubic Polynomial 

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In this section, we will discuss Factorization of Cubic Polynomial.

The degree of the cubic (highest exponent) polynomial is 3.
There are different identities in Factorization of Cubic Polynomial .
They are as follows :
1) a³ + b³ + 3a²b + 3b²a = (a + b)³

2) a³ - b³ - 3a²b + 3b²a = (a - b)³

3) a³ + b³ = (a + b)(a² - ab + b²)

4) a³- b³ = (a - b)(a² + ab + b²)

5) a³ + b³+ c³- 3abc
= (a + b + c)(a² + b²+ c² - ab - bc - ac)

6) If a + b + c = 0 then a³ + b³+ c³=3abc

Using Identities

Examples :

1)27x³- 8

Solution :
As first term and the 2nd term are perfect cube, we use the identity of cube

(a3 - b3) = ( a - b)(a2 +ab + b2)

a = 3x and b = 2

27x³- 8 = (3x -2)[(3x)² + 6x + 2²)]

= (3x -2)(9x² + 6x + 4)are the factors.

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2) 8x³ +1

Solution :
As first term and the 2nd term are perfect cube, we use the identity of cube

(a3 + b3) = ( a + b)(a2 -ab + b2)

a = 2x and b = 1

8x³+ 1 = (2x + 1)[(2x)² - 2x + 1²)]

= (2x +1 )(4x² - 2x + 1)are the factors.

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3) 27x³ + y³ + 9x²y + 3xy²

Solution :
As in the given equation there are two perfect cubes, so we can use 1st identity.

27x³ + y³ + 27x²y + 9xy²

= (3x)³ + (y)³ + 3(3x)²y + 3(3x)y²

= (3x + y)³ [ here a = 3x and b= y;a³ + b³ + 3a²b + 3b²a = (a + b)³ ]

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4) (1.5)³ - (0.9)³ - (0.6)³

Solution :
(1.5)³ - (0.9)³ + (-0.6)³

Let a = 1.5, b = - 0.9 and c = -0.6

As a + b + c = 1.5 + 0.9 - 0.6 = 0

a³ + b³+ c³ = 3abc

= 3(1.5)(- 0.9)(-0.6)

= 2.430

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Factoring

• Factorization by common factor
• Factorization by Grouping
• Factorization using Identities
• Factorization of Cubic Polynomial
• Solved Examples on Factorization

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