FACTORING SUM AND DIFFERENCE OF CUBES

The sum or difference of two cubes can be factored into a product of a binomial times a trinomial.

That is, 

a³ - b³  =  (a - b)(a² + ab + b²)

a³ + b³  =  (a + b)(a² + ab + b²)

When we have an expression like a³ - b³ or a³ + b³, we can write it as product of a binomial and a trinomial. 

Sum of Two Cubes

In sum of two cubes, the binomial factor on the right side of the equation has a middle sign that is positive. And also, the middle sign of the trinomial factor is always opposite the middle sign of the given problem. Therefore, it is negative.

Difference of Two Cubes

In difference of two cubes, the binomial factor on the right side of the equation has a middle sign that is negative. And also, the middle sign of the trinomial factor is always opposite the middle sign of the given problem. Therefore, it is positive.

Example 1 :

Factor : 

m³ + 8

Solution :

m³ + 8  =  m³ + 2³

=  m³ + 2³

=  (m + 2)(m² - 2m + 2²)

=  (m + 2)(m² - 2m + 4)

Example 2 :

Factor : 

a³ - 125

Solution :

a³ - 125  =  a³ - 5³

=  a³ - 5³

=  (a - 5)(a² + 5a + 5²)

=  (a - 5)(a² + 5a + 25)

Example 3 :

Factor : 

x³ + 8y³

Solution :

x³ + 8y³  =  x³ + 2³y³

=  x³ + (2y)³

=  (x + 2y)[x² - (x)(2y) + (2y)²]

=  (x + 2y)(x² - 2xy + 4y²)

Example 4 :

Factor : 

8x³ - 125y³

Solution :

8x³ - 125y³  =  2³x³ - 5³y³

=  (2x)³ - (5y)³

=  (2x - 5y)[(2x)² + (2x)(5y) + (5y)²]

=  (2x - 5y)(4x² + 10xy + 25y²)

Example 5 :

Factor : 

27x³ + 64y³

Solution :

27x³ + 64y³  =  3³x³ + 4³y³

=  (3x)³ + (4y)³

=  (3x + 4y)[(3x)² - (3x)(4y) + (4y)²]

=  (3x + 4y)(9x² - 12xy + 16y²)

Example 6 :

Factor : 

2m³ - 54n³

Solution :

2m³ - 54n³  =  2(m³ - 27n³)

=  2(m³ - 3³n³)

=  2[m³ - (3n)³]

=  2(m - 3n)[m² + (m)(3n) + (3n)²]

=  2(m - 3n)(m² + 3mn + 9n²)

Simplify each of the following :

Example 7 :

(x + 3)³ + (x – 3)³

Solution :

a³ + b³  =  (a + b)(a² - ab + b²)

Comparing the given question with a³ + b³, we know that 

a = x + 3 and b = x - 3

Applying these values in the formula

=  (x + 3 + x - 3)[ (x + 3)² - (x + 3)(x - 3) + (x - 3)²]

= 2x [ x² + 6x + 9 - (x² - 3²) + x² - 6x + 9]

= 2x [ x² + 9 - x² + 9 + x² + 9]

= 2x ( x² + 27)

= 2x³ + 54x

Example 8 :

If a + b = 10 and ab = 21, find the value of a³ + b³

Solution :

a³ + b³  =  (a + b)(a² - ab + b²)

By applying these values in the formula, we get

= 10(a² - 21 + b²) -----(1)

a² + b² = (a + b)² - 2ab

= (10)² - 2(21)

= 100 - 42

a² + b² = 58

Apply this value in (1)

= 10(58 - 21)

= 10(37)

= 370

Example 9 :

If a – b = 4 and ab = 21, find the value of a³ – b³.

Solution :

a³ - b³  =  (a - b)(a² + ab + b²)

By applying these values in the formula, we get

= 4(a² + 21 + b²) -----(1)

a² + b² = (a - b)² + 2ab

= (4)² + 2(21)

= 16 + 42

a² + b² = 58

Apply this value in (1)

= 4(58 + 21)

= 4(79)

= 316

Example 10 :

If x + 1/x = 5, find the value of x³ + 1/x³ .

Solution :

a³ + b³  =  (a + b)(a² - ab + b²)

x³ + 1/x³  =  (x + 1/x)(x² - x(1/x) + (1/x)²)

=  (x + 1/x)(x² - 1 + (1/x)²) ----(1)

By applying these values in the formula, we get

a² + b² = (a + b)² - 2ab

x² + 1/x² =  (x + 1/x)² - 2x(1/x)

=  (x + 1/x)² - 2

= 5² - 2

x² + 1/x² = 23

apply this value in (1), we get

=  5(23 - 1)

= 5(22)

= 110

Example 11 :

If x - 1/x = 7, find the value of x³ - 1/x³ .

Solution :

a³ - b³  =  (a - b)³ - 3ab(a - b)

x³ - 1/x³  =  (x - 1/x)³ - 3x(1/x)(x - 1/x)

=  7³ - 3(7)

= 343 - 21

= 322

Example 12 :

If (x² + 1/x²) = 98, find the value of x³ + 1/x³.

Solution :

a² + b² = (a + b)² - 2ab

x² + 1/x² = (x + 1/x)² - 2x(1/x)

98 = (x + 1/x)² - 2

(x + 1/x)² = 98 + 2

(x + 1/x)² = 100

x + 1/x = 10

a³ + b³  =  (a + b)³ - 3ab(a + b)

x³ + 1/x³  =  (x + 1/x)³ - 3x(1/x)(x + 1/x)

=  10³ - 3(10)

= 1000 - 30

= 970

Example 13 :

If 2x + 3y = 13 and xy = 6, find the value of 8x³ + 27y³.

Solution :

a³ + b³  =  (a + b)³ - 3ab(a + b)

8x³ + 27y³ = (2x)³ + (3y)³

= (2x + 3y)³ - 3(2x)(3y)(2x + 3y)

= (13)³ - 18xy(13)

= (13)³ - 18(6)(13)

= 2197 - 1404

8x³ + 27y³ = 793

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