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.

7th Grade Math Problems

8th Grade Math Practice

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