FORMULA FOR A MINUS B THE WHOLE CUBE

Expand and simplify :

Problem 1 :

(x - 1)³

Solution :

(x - 1)³ is in the form of (a - b)³

Comparing (a - b)³ and (x - 1)³ , we get

a = x, b = 1

Substitute x for a and 1 for b

(x - 1)³ = x³ - 3(x²)(1) + 3(x)(1²) - 1³

(x - 1)³ = x³ - 3x² + 3x - 1

Problem 2 :

(x - 5)³

Solution :

(x - 5)³ is in the form of (a - b)³

Comparing (a - b)³ and (x - 5)³ , we get

a = x, b = 5

(x - 5)³ = x³ - 3(x²)(5) + 3(x)(5²) – (5)³

(x - 5)³ = x³ - 15x² + 75x - 125

Problem 3 :

(x - 4)³

Solution :

Here a = x and b = 4

(x - 4)³ = x³ - 3(x²)(4) + 3(x)(4²) – (4)³

(x - 4)³ = x³ - 12x² + 48x - 64

Problem 4 :

(x - y)³

Solution :

Here a = x and b = y

(x - y)³ = x³ - 3(x²)(y) + 3(x)(y²) – (y)³

(x - y)³ = x³ - 3x²y + 3xy² - y³

Problem 5 :

(2 - y)³

Solution :

Here a = 2 and b = y

(2 - y)³ = (2)³ - 3(2²)(y) + 3(2)(y²) – (y)³

(2 - y)³ = 8 - 12y + 6y² - y³

Problem 6 :

(2x - 1)³

Solution :

Here a = 2x and b = 1

(2x - 1)³ = (2x)³ - 3(2x)²(1) + 3(2x)(1²) - 1³

(2x - 1)³ = 8x³ - 12x² + 6x - 1

Problem 7 :

(3x - 1)³

Solution :

Here a = 3x and b = 1

(3x - 1)³ = (3x)³ - 3(3x)²(1) + 3(3x)(1²) - 1³

(3x - 1)³ = 27x³ - 27x² + 9x - 1

Problem 8 :

(2y – 3x)³

Solution :

Here a = 3x and b = 2y

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

(3x - 2y)³ = 27x³ - 54x²y + 36xy² - 8y³