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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