Cube of a Binomial

How do you get the cube of a binomial?

For cubing a binomial we need to know the formulas for the sum of cubes and the difference of cubes.

Sum of cubes:

The sum of a cubed of two binomial is equal to the cube of the first term, plus three times the square of the first term by the second term, plus three times the first term by the square of the second term, plus the cube of the second term.

(a + b)3 = a3 + 3a2b + 3ab2 + b3

            = a3 + 3ab (a + b) + b3

Difference of cubes:     

The difference of a cubed of two binomial is equal to the cube of the first term, minus three times the square of the first term by the second term, plus three times the first term by the square of the second term, minus the cube of the second term.

(a – b)3 = a3 – 3a2b + 3ab2 – b3

            = a3 – 3ab (a – b) – b3


Worked-out examples for the expansion of cube of a binomial:

Simplify the following by cubing:

1. (x + 5y)3 + (x – 5y)3

Solution:

We know, (a + b)3 = a3 + 3a2b + 3ab2 + b3

and,

(a – b)3 = a3 – 3a2b + 3ab2 – b3

Here, a = x and b = 5y

Now using the formulas for cube of two binomials we get,

= x3 + 3.x2.5y + 3.x.(5y)2 + (5y)3 + x3 - 3.x2.5y + 3.x.(5y)2 - (5y)3

= x3 + 15x2y + 75xy2 + 125 y3 + x3 - 15x2y + 75xy2 - 125 y3

= 2x3 + 150xy2

Therefore, (x + 5y)3 + (x – 5y)3 = 2x3 + 150xy2


2. \((\frac{1}{2} x + \frac{3}{2} y)^{3} + (\frac{1}{2} x - \frac{3}{2} y)^{3}\)

Solution:

Here a = \(\frac{1}{2} x, b = \frac{3}{2} y\)

 \(=(\frac{1}{2} x)^{3} + 3\cdot (\frac{1}{2} x)^{2} \cdot  \frac{3}{2} y + 3 \cdot \frac{1}{2} x \cdot (\frac{3}{2}y)^{2} + (\frac{3}{2}y)^{3} + (\frac{1}{2} x)^{3} - 3\cdot (\frac{1}{2} x)^{2} \cdot  \frac{3}{2} y + 3 \cdot \frac{1}{2} x \cdot (\frac{3}{2}y)^{2} - (\frac{3}{2}y)^{3}\)

 \(=\frac{1}{8} x^{3} + \frac{9}{8} x^{2} y + \frac{27}{8} x y^{2} + \frac{27}{8} y^{3} + \frac{1}{8} x^{3} - \frac{9}{8} x^{2} y + \frac{27}{8} x y^{2} - \frac{27}{8} y^{3}\)

 \(=\frac{1}{8} x^{3} + \frac{1}{8} x^{3} + \frac{27}{8} x y^{2} + \frac{27}{8} x y^{2}\)

 \(=\frac{1}{4} x^{3} + \frac{27}{4} x y^{2} \)

Therefore, \[(\frac{1}{2} x + \frac{3}{2} y)^{3} + (\frac{1}{2} x - \frac{3}{2} y)^{3} = \frac{1}{4} x^{3} + \frac{27}{4} x y^{2} \]


3. (2 – 3x)3 – (5 + 3x)3

Solution:

(2 – 3x)3 – (5 + 3x)3

= {23 - 3.22.(3x) + 3.2.(3x)2 - (3x)3} – {53 + 3.52.(3x) + 3.5.(3x)2 + (3x)3}

= {8 – 36x + 54 x2 - 27 x3} – {125 + 225x + 135x2 + 27 x3}

= 8 – 36x + 54 x2 - 27 x3 – 125 - 225x - 135x2 - 27 x3

= 8 – 125 – 36x - 225x + 54 x2 - 135x2 - 27 x3 - 27 x3

= -117 – 261x - 81 x2 - 54 x3

Therefore, (2 – 3x)3 – (5 + 3x)3 = -117 – 261x - 81 x2 - 54 x3


4. (5m + 2n)3 - (5m – 2n)3

Solution:

(5m + 2n)3 - (5m – 2n)3

= {(5m)3 + 3.(5m)2. (2n) + 3. (5m). (2n)2 + (2n)3} – {(5m)3 - 3.(5m)2. (2n) + 3. (5m). (2n)2 - (2n)3}

= {125 m3 + 150 m2 n + 60 m n2 + 8 n3} – {125 m3 - 150 m2 n + 60 m n2 - 8 n3}

= 125 m3 + 150 m2 n + 60 m n2 + 8 n3 – 125 m3 + 150 m2 n - 60 m n2 + 8 n3

= 125 m3 – 125 m3 + 150 m2 n + 150 m2 n + 60 m n2 - 60 m n2 + 8 n3 + 8 n3

= 300 m2 n + 16 n3

Therefore, (5m + 2n)3 - (5m – 2n)3 = 300 m2 n + 16 n3


The steps to find the mixed problem on cube of a binomial will help us to expand the sum or difference of two cubes.







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