Cube of the Sum of Two Binomials

What is the formula for the cube of the sum of two binomials?

To determine cube of a number means multiplying a number with itself three times similarly, cube of a binomial means multiplying a binomial with itself three times.

(a + b) (a + b) (a + b) = (a + b)³

or, (a + b) (a + b) (a + b) = (a + b) (a + b)²

                                    = (a + b) (a² + 2ab + b²),
                                    [Using the formula of (a + b)² = a² + 2ab + b²]

                                    = a(a² +2ab + b²) + b(a² + 2ab + b²)

                                    = a³ + 2a² b + ab² + ba² + 2ab² + b³

                                    = a³ + 3a² b + 3ab² + b³







Therefore, (a + b)³ = a³ + 3a² b + 3ab² + b³

Thus, we can write it as; a = first term, b = second term

(First term + Second term)³ = (first term)³ + 3 (first term)² (second term) + 3 (first term) (second term)² + (second term)³

So, the formula for the cube of the sum of two terms is written as:

(a + b)³ = a³ + 3a²b + 3ab² + b³

            = a³ + b³ + 3ab (a + b)

Worked-out examples to find the cube of the sum of two binomials:

1. Determine the expansion of (3x - 2y)³

Solution:

We know, (a + b)³ = a³ + 3a² b + 3ab² + b³

(3x - 2y)³

Here, a = 3x, b = 2y

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

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

= 27x³ + 54x²y + 36xy² + 8y³

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


2. Use the formula and evaluate (105)³.

Solution:

(105)³

= (100 + 5)³

We know, (a + b)³ = a³ + 3a² b + 3ab² + b³

Here, a = 100, b = 5

= (100)³ + 3 (100)² (5) + 3 (100) (5)² + (5)³

= 1000000 + 15 (10000) + 300 (25) + 125

= 1000000 + 150000 + 7500 + 125

= 1157625

Therefore, (105)³ = 1157625








3. Find the value of x³ + 27y³ if x + 3y = 5 and xy = 2.

Solution:

Given, x + 3y = 5

Now cube both sides we get,

(x + 3y)³ = (5)³

We know, (a + b)³ = a³ + 3a² b + 3ab² + b³

Here, a = x, b = 3y

⇒ x³ + 3 (x)² (3y) + 3 (x)(3y)² + (3y)³ = 343

⇒ x³ + 9(x)² y + 27xy² 27y³ = 343

⇒ x³ + 9xy [x + 3y] + 27y³ = 343

Substituting the value of x + 3y = 5 and xy = 2, we get

⇒ x³ + 9 (2) (5) + 27y³ = 343

⇒ x³ + 90 + 27y³ = 343

⇒ x³ + 27y³ = 343 – 90

⇒ x³ +27y³ = 253

Therefore, x³ + 27y³ = 253

4. If x - \(\frac{1}{x}\)= 5, find the value of \(x^{3}\) - \(\frac{1}{x^{3}}\)

Solution:

x - \(\frac{1}{x}\) = 5

Cubing both sides, we get

 (x -  \(\frac{1}{x}\))\(^{3}\) =  \(5^{3}\)

\(x^{3}\) –  3 (x) (\(\frac{1}{x}\)) [ x - \(\frac{1}{x}\)] – (\(\frac{1}{x}\))\(^{3}\) = 216

\(x^{3}\)  – 3 (x - \(\frac{1}{x}\)) – \(\frac{1}{x^{3}}\) = 216                       

\(x^{3}\) –  \(\frac{1}{x^{3}}\) – 3 (x - \(\frac{1}{x}\)) = 216

\(x^{3}\) –  \(\frac{1}{x^{3}}\) – 3 × 5 = 216, [Putting the value of  x - \(\frac{1}{x}\)= 5]                  

\(x^{3}\) –  \(\frac{1}{x^{3}}\) – 15 = 216

\(x^{3}\) – \(\frac{1}{x^{3}}\) = 216 + 15                               

\(x^{3}\) – \(\frac{1}{x^{3}}\) = 231

Thus, to expand the cube of the sum of two binomials we can use the formula to evaluate.

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