FORMULA FOR A PLUS B THE WHOLE CUBE

Use the binomial expansion for (a + b)³ to expand and simplify:

Problem 1 :

(x + 1)³

Solution :

By comparing (a + b)³ and (x + 1)³, we get

a = x, b = 1

(a + b)³ = a³ + 3a²b + 3ab² + b³

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 + 3)³

Solution :

By comparing (a + b)³ and (x + 3)³, we get

a = x, b = 3

(a + b)³ = a³ + 3a²b + 3ab² + b³

Substitute x for a and 3 for b

(x + 3)³ = x³ + 3(x²) (3) + 3(x) (3²) + 3³

(x + 3)³ = x³ + 9x² + 27x + 27

Problem 3 :

(x + 4)³

Solution :

By comparing (a + b)³ and (x + 4)³, we get

a = x, b = 4

(a + b)³ = a³ + 3a²b + 3ab² + b³

Substitute x for a and 4 for b

(x + 4)³ = x³ + 3(x²) (4) + 3(x) (4²) + 4³

(x + 4)³ = x³ + 12x² + 48x + 64

Problem 4 :

(2 + y)³

Solution :

By comparing (a + b)³ and (2 + y)³, we get

a = 2, b = y

(a + b)³ = a³ + 3a²b + 3ab² + b³

Substitute 2 for a and y for b

(2 + y)³ = 2³ + 3(2²) (y) + 3(2) (y²) + y³

(2 + y)³ = 8 + 12y + 6y² + y³

Problem 5 :

(2x + 1)³

Solution :

By comparing (a + b)³ and (2x + 1)³, we get

a = 2x, b = 1

(a + b)³ = a³ + 3a²b + 3ab² + b³

Substitute 2x for a and 1 for b

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

(2x + 1)³ = 8x³ + 12x² + 6x + 1

Problem 6 :

(2y + 3x)³

Solution :

(2y + 3x)³ is changed to (3x + 2y)³ 

By comparing (a + b)³ and (3x + 2y)³, we get

a = 3x, b = 2y

(a + b)³ = a³ + 3a²b + 3ab² + b³

Substitute 3x for a and 2y for b

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

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

Problem 7 :

Find the polynomial that represent the volume of a cube if each side measures x + 4?

Solution :

Volume of a cube V = s³

Here, s is the side length of the cube

V = (x + 4)³

a = x, b = 4

(a + b)³ = a³ + 3a²b + 3ab² + b³

Substitute x for a and 4 for b

(x + 4)³ = x³ + 3(x²) (4) + 3(x) (4)² + (4)³

(x + 4)³ = x³ + 12x² + 48x + 64