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

Problem 8 :

If 3x + 4y = 11 and xy = 2, find the value of 27x³ + 64y³

Solution :

3x + 4y = 11 and xy = 2

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a + b)³ = a³ + b³ + 3ab(a + b)

(3x + 4y)³ = (3x)³ + (4y)³ + 3(3x)(4y)(3x + 4y)

(3x + 4y)³ = (3x)³ + (4y)³ + 36xy(3x + 4y)

Applying the givne values, we get

11³ = 27x³ + 64y³ + 36(2)(11)

1331 = 27x³ + 64y³ + 792

27x³ + 64y³ = 1331 - 792

27x³ + 64y³ = 539

Problem 9 :

If x² + 1/x² = 83, find the value of x³ - 1/x³

Solution :

x² + 1/x² = 83

a² + b² = (a - b)² + 2ab

Here a = x and b = 1/x

x² + 1/x² = (x - 1/x)² + 2x(1/x)

83 = (x - 1/x)² + 2

(x - 1/x)² = 83 - 2

(x - 1/x)² = 81

(x - 1/x) = √81

(x - 1/x) = 9

(a - b)³ = a³ - 3a²b + 3ab² - b³

(a - b)³ = a³ - b³ - 3ab(a - b)

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

(x - 1/x)³ = x³ - 1/x³ - 3 (x - 1/x)

Applying the value of (x + 1/x) = √85, we get

9³ = x³ - 1/x³ - 3 (9)

729 = x³ - 1/x³ - 27

 x³ - 1/x³ = 729 + 27

 x³ - 1/x³ = 756

Problem 10 :

The vaue 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 / 9(x - y)(y - z)(z - x)

(a + b + c)³ = a³ + b³ + c³ + 3 (a +b) (b + c) (a+ c)

When a + b + c = 0

a³ + b³ + c³  = - 3 (a +b) (b + c) (a+ c)

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

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

= 0

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

= - 3 (x - z) (y - x) (z - y)

Applying the value, we get

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

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

= 3/9

= 1/3

So, option c is correct.