A CUBE MINUS B CUBE FORMULA

In this section, we are going to see the formula  for

a³ - b³

We already know the formula/expansion for (a - b)³.

That is, 

(a - b)³  =  a³ - b³ - 3ab(a - b)

Case 1 : 

(a - b)³  =  a³ - b³ - 3ab(a - b)

Add 3ab(a - b) to each side. 

(a - b)³ + 3ab(a - b)  =  a³ - b³

Therefore, the formula for (a³ - b³) is 

a³ - b³  =  (a - b)³ + 3ab(a - b)

Case 2 : 

From case 1, 

a³ - b³  =  (a - b)³ + 3ab(a - b)

a³ - b³  =  (a - b)[(a - b)² + 3ab]

a³ - b³  =  (a - b)[a² - 2ab + b² + 3ab]

a³ - b³  =  (a - b)(a² + ab + b²)

Therefore, the formula for (a³ - b³) is 

a³ - b³  =  (a - b)(a² + ab + b²)

So, 

(a - b) and (a² + ab + b²)

are the factors of (a³ - b³).

Note : 

Based on our need, either we can use the formula in case 1 or in case 2 for (a³ - b³).

Practice Questions

Question 1 :

Factor :

x³ - 1

Solution :

Write (x³ - 1) in the form of (a³ - b³).

x³ - 1  =  x³ - 1³

(x³ - 1³) is in the form of (a³ - b³).

Comparing (a³ - b³) and (x³ - 1³), we get

a  =  x

b  =  1

Write the formula for (a³ - b³) given in case 2 above.

a³ - b³  =  (a - b)(a² + ab + b²)

Substitute x for a and 1 for b. 

x³ - 1³  =  (x - 2)(x² + x(1) + 1²)

x³ + 1  =  (x - 1)(x² + x + 1)

Question 2 :

Factor :

8x³ - 27y³

Solution :

Write (8x³ - 27y³) in the form of (a³ - b³).

8x³ - 27y³  =  (2x)³ - (3y)³

(2x)³ - (3y)³ is in the form of (a³ - b³).

Comparing (a³ - b³) and (2x)³ - (3y)³, we get

a  =  2x

b  =  3y

Write the formula for (a³ - b³) given in case 2 above.

a³ - b³  =  (a - b)(a² + ab + b²)

Substitute 2x for a and 3y for b. 

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

8x³ - 27y³  =  (2x - 3y)(4x² + 6xy + 9y²)

Question 3 :

Factor :

125p³ - 64q³

Solution :

Write (125p³ - 64q³) in the form of (a³ - b³).

125p³ - 64q³  =  (5p)³ - (4q)³

(5p)³ - (4q)³ is in the form of (a³ - b³).

Comparing (a³ - b³) and (5p)³ - (4q)³, we get

a  =  5p

b  =  4q

Write the formula for (a³ - b³) given in case 2 above.

a³ - b³  =  (a - b)(a² + ab + b²)

Substitute 5p for a and 4q for b.  9 

(5p)³ - (4q)³  =  (5p - 4q)[(5p)² + (5p)(4q) + (4q)²]

125p³ - 64q³  =  (5p - 4q)(25p² + 20pq + 16q²)

Question 4 :

Find the value of (m³ - n³), if m - n = 3 and mn = 28.

Solution :

Write (m³ - n³) in terms of (m - n) and mn using the formula given in case 1 above.

m³ - n³  =  (m - n)³ + 3mn(m - n)

Substitute 3 for (m - n) and 28 for xy. 

x³ - y³  =  (3)³ + 3(28)(3)

x³ - y³  =  27 + 252

x³ - y³  =  279

Question 5 :

The value of

(x - y)³ + (y - z)³ + (z - x)³ / 9(x - y)(y - z) (z - x)

is equal to

a) 0    b) 1/9    c)  1/3    d)  1

Solution :

(x - y)³ + (y - z)³ + (z - x)³

Let a = x - y, b = y - z and c = z - x

a + b + c = x - y + y - z + z - x

a + b + c = 0

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

When a + b + c = 0

a³ + b³ + c³ - 3abc = 0

a³ + b³ + c³ = 3abc

(x - y)³ + (y - z)³ + (z - x)³ = 3(x - y)(y - z)(z - x)

By applying this in the given expression, we get

= 3(x - y)(y - z)(z - x) / 9(x - y)(y - z) (z - x)

= 3/9

= 1/3

so, option c is correct.

Question 6 :

Simplify the following :

a) 9(3a³ - 24b³) / (9a² - 36b²)

b) (s³ + 125 t³) / (s² - 2st - 35t²)

Solution :

a) 9(3a³ - 24b³) / (9a² - 36b²)

9(3a³ - 24b³) = 9(3) (a³ - 8b³)

= 27(a³ - 2³b³)

= 27(a³ - (2b)³)

= 27 (a - 2b)(a² + a(2b) + (2b)²)

= 27 (a - 2b)(a² + 2ab + 4b²) ------(1)

(9a² - 36b²) = 9(a² - 4b²)

= 9(a² - 2²b²)

= 9(a² - (2b)²)

= 9(a + 2b)(a - 2b) ------(2)

(1) / (2)

= 27 (a - 2b)(a² + 2ab + 4b²) /  9(a + 2b)(a - 2b)

= 3(a² + 2ab + 4b²) / (a + 2b)

b) (s³ + 125 t³) / (s² - 2st - 35t²)

(s³ + 125 t³) = (s³ + 5³ t³)

= s³ + (5t)³

= (s + 5t)(s² - s(5t) + (5t)²)

= (s + 5t)(s² + 5st + 25t²) -----(1)

= (s² - 2st - 35t²) ----(2)

(1) / (2)

= (s + 5t)(s² + 5st + 25t²) / (s² - 2st - 35t²)

Question 7 :

If a + b + c = 5 and ab + bc + ca = 10, then prove that a³ + b³ + c³ – 3abc = – 25.

Solution :

a + b + c = 5 and ab + bc + ca = 10,

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

a³ + b³ + c³ - 3abc = 5(a² + b² + c² - (ab + bc + ca))

= 5(a² + b² + c² - 10) ---(1)

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

5² = a² + b² + c² + 2(ab + bc + ca)

25 = a² + b² + c² + 2(10)

25 = a² + b² + c² + 20

a² + b² + c² = 25 - 20

a² + b² + c² = 5

Applying the above value of in (1), we get

= 5(5 - 10)

= 5(-5)

a³ + b³ + c³ - 3abc = -25

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.