PAS we know,(x + 1/x)3 = x3 + 1/x3 + 3 (x + 1/x)p3 = k + 3pk = p3 - 3pk/p = p2 - 3Short TRICK:\({\left( {{x^3} + \frac{1}{{{x^3}}} - k} \right)^2} + {\left( {x + \frac{1}{x} - p} \right)^2} = 0\)Put x = 1, then(2 - k)2 + (2 - p)2 = 0k = 2 and p = 2k/p = 2/2 = 1From OPTION 4.p2 - 322 - 34 - 31 (satisfied)

"> PAS we know,(x + 1/x)3 = x3 + 1/x3 + 3 (x + 1/x)p3 = k + 3pk = p3 - 3pk/p = p2 - 3Short TRICK:\({\left( {{x^3} + \frac{1}{{{x^3}}} - k} \right)^2} + {\left( {x + \frac{1}{x} - p} \right)^2} = 0\)Put x = 1, then(2 - k)2 + (2 - p)2 = 0k = 2 and p = 2k/p = 2/2 = 1From OPTION 4.p2 - 322 - 34 - 31 (satisfied)

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If \({\left( {{x^3} + \frac{1}{{{x^3}}} - k} \right)^2} + {\left( {x + \frac{1}{x} - p} \right)^2} = 0,\) where k and p are real numbers and x ≠ 0, then k/p is equal to:

General Aptitude Algebra in General Aptitude 9 months ago

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As we know,If, (a - b)2 + (c - d)2 = 0a = b and c = d\({\left( {{x^3} + \frac{1}{{{x^3}}} - k} \right)^2} + {\left( {x + \frac{1}{x} - p} \right)^2} = 0\)\({\left( {{x^3} + \frac{1}{{{x^3}}} - k} \right)^2} = 0\)x3 + 1/x3 = kAnd\({\left( {x + \frac{1}{x} - p} \right)^2} = 0\)x + 1/x = PAS we know,(x + 1/x)3 = x3 + 1/x3 + 3 (x + 1/x)p3 = k + 3pk = p3 - 3pk/p = p2 - 3Short TRICK:\({\left( {{x^3} + \frac{1}{{{x^3}}} - k} \right)^2} + {\left( {x + \frac{1}{x} - p} \right)^2} = 0\)Put x = 1, then(2 - k)2 + (2 - p)2 = 0k = 2 and p = 2k/p = 2/2 = 1From OPTION 4.p2 - 322 - 34 - 31 (satisfied)

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Posted on 03 Nov 2024, this text provides information on General Aptitude related to Algebra in General Aptitude. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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