CUBE OF A BINOMIAL

Example 1 :

Expand (a - b)³.

Solution :

We know that

(a + b)³ = a³ + 3a²b + 3ab² + b³

Replace 'b' by '-b'.

(a + (-b))³ = a³ + 3a²(-b) + 3a(-b)² + (-b)³

(a - b)³ = a³ - 3a²b + 3ab² - b³

or

(a - b)³ = a³ - 3ab(a - b) - b³

Example 2 :

Expand (y + 5)³.

Solution :

We know that

(a + b)³ = a³ + 3a²b + 3ab² + b³

Substitute a = y, and b = 5.

(y + 5)³ = y³ + 3y²(5) + 3y(5)² + 5³

= y³ + 15y² + 3y(25) + 125

= y³ + 15y² + 75y + 125

Example 3 :

Expand (x - 7)³.

Solution :

We know that

(a + b)³ = a³ + 3a²b + 3ab² + b³

Substitute a = x, and b = -7.

(x - 7)³ = x³ + 3x²(-7) + 3x(-7)² + (-7)³

= x³ - 21x² + 3x(49) - 343

= x³ - 21x² + 147x - 343

Example 4 :

Expand (5x + 3y)³.

Solution :

We know that

(a + b)³ = a³ + 3a²b + 3ab² + b³

Substitute a = 5x, and b = 3y.

(5x + 3y)³ = (5x)³ + 3(5x)²(3y) + 3(5x)(3y)² + (3y)³

= 125x³ + 3(25x²)(3y) + 3(5x)(9y²) + 27y³

= 125x³ + 225x²y + 135xy² + 27y³

Example 5 :

Expand (3p - 4q)³.

Solution :

We know that

(a + b)³ = a³ + 3a²b + 3ab² + b³

Substitute a = 3p, and b = -4q.

(3p - 4q)³ = (3p)³ + 3(3p)²(-4q) + 3(3p)(-4q)² + (-4q)³

= 27p³ + 3(9p²)(-4q) + 3(3p)(16q²) + (-64q³)

= 27p³ - 108p²q + 144pq² - 64q³

Example 6 :

Expand (x + 1/y)³.

Solution :

We know that

(a + b)³ = a³ + 3a²b + 3ab² + b³

Substitute a = x, and b = 1/y.

(x + 1/y)³ = x³ + 3(x)²(1/y) + 3(x)(1/y)² + (1/y)³

= x³ + 3x²/y + 3x/y² + 1/y³

Example 7 :

Evaluate using identity : 98³.

Solution :

98³ = (100 - 2)³

We know that

(a + b)³ = a³ + 3a²b + 3ab² + b³

Substitute a = 100, and b = -2.

(100 - 2)³ = 100³ + 3(100)²(-2) + 3(100)(-2)² + (-2)³

98³ = 1000000 + 3(10000)(-2) + 3(100)(4) - 8

= 1000000 - 60000 + 1200 - 8

= 941192

Example 8 :

Evaluate using identity : 1001³.

Solution :

1001³ = (1000 + 1)³

We know that

(a + b)³ = a³ + 3a²b + 3ab² + b³

Substitute a = 1000, and b = 1.

(1000 + 1)³ = 1000³ + 3(1000)²(1) + 3(1000)(1)² + (1)³

(1001)³ = 1000³ + 3(1000)²(1) + 3(1000)(1)² + (1)³

= 1000000000 + 3(1000000)(1) + 3(1000)(1) + 1

= 1000000000 + 3000000 + 3000 + 1

= 1003003001

Example 9 :

Find 27x³ + 64y³, if 3x + 4y = 10 and xy = 2.

Solution :

We know that

a³ + b³ = (a + b)³ - 3ab(a + b)

Substitute a = 3x, and b = 4y.

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

27x³ + 64y³ = (3x + 4y)³ - 36(xy)(3x + 4y)

Substitute 3x + 4y = 10 and xy = 2.

27x³ + 64y³ = (10)³ - 36(2)(10)

= 1000 - 720

= 280

Example 10 :

Find x³ - y³, if x - y = 5 and xy = 14.

Solution :

We know that

a³ - b³ = (a - b)³ + 3ab(a - b)

Substitute a = x, and b = y.

x³ - y³ = (x - y)³ + 3xy(x - y)

Substitute x - y = 5 and xy = 14.

x³ - y³ = 5³ + 3(14)(5)

= 125 + 210

= 335

Example 11 :

If a + 1/a = 6, then find the value of a³ + 1/a³.

Solution :

We know that

a³ + b³ = (a + b)³ - 3ab(a + b)

Substitute a = a, and b = 1/a.

a³ + (1/a)³ = (a + 1/a)³ - 3a(1/a)(a + 1/a)

a³ + 1/a³ = (a + 1/a)³ - 3(a + 1/a)

Substitute a + 1/a = 6.

a³ + 1/a³ = (6)³ + 3(6)

= 216 + 18

= 198

Example 12 :

If (y - 1/y)³ = 27, then find the value of y³ - 1/y³.

Solution :

(y - 1/y)³ = 27

(y - 1/y)³ = 3³

y - 1/y = 3

We know that

a³ - b³ = (a - b)³ + 3ab(a - b)

Substitute a = y, and b = 1/y.

y³ - (1/y)³ = (y - 1/y)³ + 3y(1/y)(y - 1/y)

y³ - 1/y³ = (y - 1/y)³ + 3(y - 1/y)

Substitute (y - 1/y)³ = 27 and y - 1/y = 3.

y³ - 1/y³ = 27 + 3(3)

= 27 + 9

= 36

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