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Using (a–b)3 = a3 –b3 –3ab(a–b) tick the correct of these when x = p1/3 – p–1/3 (a) x3 3x = p 1/p (b) x3 3x = p – 1/p (c) x3 3x = p 1 (d) none of these?
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Using (a–b)3 = a3 –b3 –3ab(a–b) tick the correct of these when x = p1/...
Solution:
Given: x = p1/3 – p–1/3
We need to find the correct option out of (a), (b), (c), and (d) using (a–b)3 = a3 –b3 –3ab(a–b).
Expansion of (a–b)3
(a–b)3 = a3 –b3 –3ab(a–b)
On putting a = p1/3 and b = p–1/3, we get
(p1/3 – p–1/3)3 = (p1/3)3 – (p–1/3)3 – 3(p1/3)(p–1/3)(p1/3 – p–1/3)
Simplifying the above expression, we get
p + 3p1/3 p–1/3(x) = 2p1/3 + 2p–1/3
Finding the value of x3 – 3x
Multiplying the above equation by p1/3, we get
p4/3 + 3p(x) – p2/3(x) = 2p + 2p2/3
Multiplying the above equation by p–2/3, we get
p1/3 + 3(x) – (x)(p–1/3) = 2p–1/3 + 2
Adding (x)(p–1/3) to both sides, we get
p1/3 + 3(x) = (x)(p1/3 + p–1/3) + 2p–1/3 + 2
Substituting the value of x = p1/3 – p–1/3, we get
p1/3 + 3(p1/3 – p–1/3) = (p1/3 – p–1/3)(p1/3 + p–1/3) + 2p–1/3 + 2
Simplifying the above expression, we get
3p1/3 – 3p–1/3 = p + 2p–1/3 + 2
Multiplying both sides by p1/3, we get
3p – 3 = p4/3 + 2p2/3 + 2p1/3
On simplifying the above expression, we get
p4/3 – 3p1/3 – 2p–1/3 – 3 = 0
Multiplying both sides by p1/3, we get
p5/3 – 3p2/3 – 2p1/3 – 3p1/3 = 0
p5/3 – 5p2/3 – 3p1/3 = 0
Multiplying both sides by p1/3, we get
p2 + 5p – 3 = 0
Solving the above quadratic equation, we get
p = (–5 ± √37)/2
Comparing the given options
(a) x3 3x = p 1/p
On substituting the value of x in the above
Given: x = p1/3 – p–1/3
We need to find the correct option out of (a), (b), (c), and (d) using (a–b)3 = a3 –b3 –3ab(a–b).
Expansion of (a–b)3
(a–b)3 = a3 –b3 –3ab(a–b)
On putting a = p1/3 and b = p–1/3, we get
(p1/3 – p–1/3)3 = (p1/3)3 – (p–1/3)3 – 3(p1/3)(p–1/3)(p1/3 – p–1/3)
Simplifying the above expression, we get
p + 3p1/3 p–1/3(x) = 2p1/3 + 2p–1/3
Finding the value of x3 – 3x
Multiplying the above equation by p1/3, we get
p4/3 + 3p(x) – p2/3(x) = 2p + 2p2/3
Multiplying the above equation by p–2/3, we get
p1/3 + 3(x) – (x)(p–1/3) = 2p–1/3 + 2
Adding (x)(p–1/3) to both sides, we get
p1/3 + 3(x) = (x)(p1/3 + p–1/3) + 2p–1/3 + 2
Substituting the value of x = p1/3 – p–1/3, we get
p1/3 + 3(p1/3 – p–1/3) = (p1/3 – p–1/3)(p1/3 + p–1/3) + 2p–1/3 + 2
Simplifying the above expression, we get
3p1/3 – 3p–1/3 = p + 2p–1/3 + 2
Multiplying both sides by p1/3, we get
3p – 3 = p4/3 + 2p2/3 + 2p1/3
On simplifying the above expression, we get
p4/3 – 3p1/3 – 2p–1/3 – 3 = 0
Multiplying both sides by p1/3, we get
p5/3 – 3p2/3 – 2p1/3 – 3p1/3 = 0
p5/3 – 5p2/3 – 3p1/3 = 0
Multiplying both sides by p1/3, we get
p2 + 5p – 3 = 0
Solving the above quadratic equation, we get
p = (–5 ± √37)/2
Comparing the given options
(a) x3 3x = p 1/p
On substituting the value of x in the above
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Using (a–b)3 = a3 –b3 –3ab(a–b) tick the correct of these when x = p1/3 – p–1/3 (a) x3 3x = p 1/p (b) x3 3x = p – 1/p (c) x3 3x = p 1 (d) none of these?
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Using (a–b)3 = a3 –b3 –3ab(a–b) tick the correct of these when x = p1/3 – p–1/3 (a) x3 3x = p 1/p (b) x3 3x = p – 1/p (c) x3 3x = p 1 (d) none of these? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about Using (a–b)3 = a3 –b3 –3ab(a–b) tick the correct of these when x = p1/3 – p–1/3 (a) x3 3x = p 1/p (b) x3 3x = p – 1/p (c) x3 3x = p 1 (d) none of these? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Using (a–b)3 = a3 –b3 –3ab(a–b) tick the correct of these when x = p1/3 – p–1/3 (a) x3 3x = p 1/p (b) x3 3x = p – 1/p (c) x3 3x = p 1 (d) none of these?.
Using (a–b)3 = a3 –b3 –3ab(a–b) tick the correct of these when x = p1/3 – p–1/3 (a) x3 3x = p 1/p (b) x3 3x = p – 1/p (c) x3 3x = p 1 (d) none of these? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about Using (a–b)3 = a3 –b3 –3ab(a–b) tick the correct of these when x = p1/3 – p–1/3 (a) x3 3x = p 1/p (b) x3 3x = p – 1/p (c) x3 3x = p 1 (d) none of these? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Using (a–b)3 = a3 –b3 –3ab(a–b) tick the correct of these when x = p1/3 – p–1/3 (a) x3 3x = p 1/p (b) x3 3x = p – 1/p (c) x3 3x = p 1 (d) none of these?.
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