Subscribe to our YouTube channel for the latest videos, updates, and tips.
Here we will solve different types of application problems on expanding of (a ± b)\(^{3}\) and its corollaries.
1. Expanding the following:
(i) (1 + x)\(^{3}\)
(ii) (2a – 3b)\(^{3}\)
(iii) (x + \(\frac{1}{x}\))\(^{3}\)
Solution:
(i) (1 + x)\(^{3}\) = 1\(^{3}\) + 3 ∙ 1\(^{2}\) ∙ x + 3 ∙ 1 ∙ x\(^{2}\) + x\(^{3}\)
= 1 + 3x + 3x\(^{2}\) + x\(^{3}\)
(ii) (2a – 3b)\(^{3}\) = (2a)\(^{3}\) - 3 ∙ (2a)\(^{2}\) ∙ (3b) + 3 ∙ (2a) ∙ (3b)\(^{2}\) – (3b)\(^{3}\)
= 8a\(^{3}\) – 36a\(^{2}\)b + 54ab\(^{2}\) – 27b\(^{3}\)
(iii) (x + \(\frac{1}{x}\))\(^{3}\) = x\(^{3}\) + 3 ∙ x\(^{2}\) ∙ \(\frac{1}{x}\) + 3 ∙ x ∙ \(\frac{1}{x^{2}}\) + \(\frac{1}{x^{3}}\)
= x\(^{3}\) + 3x + \(\frac{3}{x}\) + \(\frac{1}{x^{3}}\).
2. Simplify: \((\frac{x}{2} + \frac{y}{3})^{3} - (\frac{x}{2} - \frac{y}{3})^{3}\)
Solution:
Given expression = \(\left \{(\frac{x}{2})^{3} + 3 \cdot (\frac{x}{2})^{2} \cdot \frac{y}{3} + 3 \cdot \frac{x}{2} \cdot (\frac{y}{3})^{2} + (\frac{y}{3})^{3}\right\} - \left \{(\frac{x}{2})^{3} - 3 \cdot (\frac{x}{2})^{2} \cdot \frac{y}{3} + 3 \cdot \frac{x}{2} \cdot (\frac{y}{3})^{2} - (\frac{y}{3})^{3}\right\}\)
= \(2\left \{3 \cdot (\frac{x}{2})^{2} \cdot \frac{y}{3} + (\frac{y}{3})^{3}\right\}\)
= \(2\left \{3 \cdot \frac{x^{2}}{4} \cdot \frac{y}{3} + \frac{y^{3}}{27}\right\}\)
= \(\frac{x^{2}y}{2} + \frac{2y^{3}}{27}\).
3. Express 8a\(^{3}\) – 36a\(^{2}\)b + 54ab\(^{2}\) – 27b\(^{3}\) as a perfect cube and find its value when a = 3, b = 2.
Solution:
Given expression = (2a)\(^{3}\) – 3(2a)\(^{2}\) ∙ 3b + 3 ∙ (2a) ∙ (3b)\(^{2}\) – (3b)\(^{3}\)
= (2a – 3b)\(^{3}\)
When a = 3 and b = 2, the value of the expression = (2 × 3 – 3 × 2)\(^{3}\)
= (6 – 6)\(^{3}\)
= (0)\(^{3}\)
= 0.
4. If x + y = 6 and x\(^{3}\) + y\(^{3}\) = 72, find xy.
Solution:
We know that (a + b)\(^{3}\) – (a\(^{3}\) + b\(^{3}\)) = 3ab(a + b).
Therefore, 3xy(x + y) = (x + y)\(^{3}\) – (x\(^{3}\) + y\(^{3}\))
Or, 3xy ∙ 6 = 6\(^{3}\) – 72
Or, 18xy = 216 – 72
Or, 18xy = 144
Or, xy = \(\frac{1}{18}\) ∙ 144
Therefore, xy = 8
5. Find a\(^{3}\) + b\(^{3}\) if a + b = 5 and ab = 6.
Solution:
We know that a\(^{3}\) + b\(^{3}\) = (a + b)\(^{3}\) - 3ab(a + b).
Therefore, a\(^{3}\) + b\(^{3}\) = 5\(^{3}\) – 3 ∙ 6 ∙ 5
= 125 – 90
= 35.
6. Find x\(^{3}\) - y\(^{3}\) if x – y = 7and xy = 2.
Solution:
We know that a\(^{3}\) - b\(^{3}\) = (a - b)\(^{3}\) + 3ab(a - b).
Therefore, x\(^{3}\) - y\(^{3}\) = (x - y)\(^{3}\) + 3xy(x - y)
= (-7)\(^{3}\) + 3 ∙ 2 ∙ (-7)
= - 343 – 42
= -385.
7. If a - \(\frac{1}{a}\) = 5, find a\(^{3}\) - \(\frac{1}{a^{3}}\).
Solution:
a\(^{3}\) - \(\frac{1}{a^{3}}\) = (a - \(\frac{1}{a}\))\(^{3}\) + 3 ∙ a ∙ \(\frac{1}{a}\)(a - \(\frac{1}{a}\))
= 5\(^{3}\) + 3 ∙ 1 ∙ 5
= 125 + 15
= 140.
8. If x\(^{2}\) + \(\frac{1}{a^{2}}\) = 7, find x\(^{3}\) + \(\frac{1}{x^{3}}\).
Solution:
We know, (x + \(\frac{1}{x}\))\(^{2}\) = x\(^{2}\) + 2 ∙ x ∙ \(\frac{1}{x}\) + \(\frac{1}{x^{2}}\)
= x\(^{2}\) + \(\frac{1}{x^{2}}\) + 2
= 7 + 2
= 9.
Therefore, x + \(\frac{1}{x}\) = \(\sqrt{9}\) = ±3.
Now, x\(^{3}\) + \(\frac{1}{x^{3}}\) = (x + \(\frac{1}{x}\))\(^{3}\) - 3 ∙ x ∙ \(\frac{1}{x}\)(x + \(\frac{1}{x}\))
= (x + \(\frac{1}{x}\))\(^{3}\) - 3(x + \(\frac{1}{x}\)).
If x + \(\frac{1}{x}\) = 3, x\(^{3}\) + \(\frac{1}{x^{3}}\) = 3\(^{3}\) - 3 ∙ 3
= 27 – 9
= 18.
If x + \(\frac{1}{x}\) = -3, x\(^{3}\) + \(\frac{1}{x^{3}}\) = (-3)\(^{3}\) - 3 ∙ (-3)
= -27 + 9
= -18.
From Problems on Expanding of (a ± b)\(^{3}\) and its Corollaries to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Jun 18, 25 02:57 PM
Jun 18, 25 01:50 AM
Jun 18, 25 01:33 AM
Jun 18, 25 01:29 AM
Jun 15, 25 04:06 PM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.