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Problems on Expanding of (a ± b)\(^{3}\) and its Corollaries
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.
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