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) (a
^{2} + 2ab + b
^{2}),
[Using the formula of (a + b)
^{2} = a
^{2} + 2ab + b
^{2}]
= a(a
^{2} +2ab + b
^{2}) + b(a
^{2} + 2ab + b
^{2})
= a
^{3} + 2a
^{2} b + ab
^{2} + ba
^{2} + 2ab
^{2} + b
^{3}
= a
^{3} + 3a
^{2} b + 3ab
^{2} + b
^{3}
Therefore, (a + b)
^{3} = a
^{3} + 3a
^{2} b + 3ab
^{2} + b
^{3}
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} = a
^{3} + 3a
^{2}b + 3ab
^{2} + b
^{3}
= a
^{3} + b
^{3} + 3ab (a + b)
Workedout 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} = a
^{3} + 3a
^{2} b + 3ab
^{2} + b
^{3}
(3x  2y)
^{3}
Here, a = 3x, b = 2y
= (3x)
^{3} + 3 (3x)
^{2} (2y) + 3 (3x)(2y)
^{2} + (2y)
^{3}
= 27x
^{3} + 3 (9x
^{2}) (2y) + 3 (3x)(4y
^{2}) + (8y
^{3})
= 27x
^{3} + 54x
^{2}y + 36xy
^{2} + 8y
^{3}
Therefore, (3x  2y)
^{3} = 27x
^{3} + 54x
^{2}y + 36xy
^{2} + 8y
^{3}
2. Use the formula and evaluate (105)
^{3}.
Solution:
(105)
^{3}
= (100 + 5)
^{3}
We know, (a + b)
^{3} = a
^{3} + 3a
^{2} b + 3ab
^{2} + b
^{3}
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 x
^{3} + 27y
^{3} 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} = a
^{3} + 3a
^{2} b + 3ab
^{2} + b
^{3}
Here, a = x, b = 3y
⇒ x
^{3} + 3 (x)
^{2} (3y) + 3 (x)(3y)
^{2} + (3y)
^{3} = 343
⇒ x
^{3} + 9(x)
^{2} y + 27xy
^{2} 27y
^{3} = 343
⇒ x
^{3} + 9xy [x + 3y] + 27y
^{3} = 343
Substituting the value of x + 3y = 5 and xy = 2, we get
⇒ x
^{3} + 9 (2) (5) + 27y
^{3} = 343
⇒ x
^{3} + 90 + 27y
^{3} = 343
⇒ x
^{3} + 27y
^{3} = 343 – 90
⇒ x
^{3} +27y
^{3} = 253
Therefore, x
^{3} + 27y
^{3} = 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.
7th Grade Math Problems
8th Grade Math Practice
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