Theory-of-Computation
November 14, 2023Question 14100 – Mathematical-Reasoning
November 14, 2023Mathematical-Reasoning
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Question 3
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The sum of a number and its inverse is –4. The sum of their cubes is:
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-52
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52
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64
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-64
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Question 3 Explanation:
Given : a + 1/a = – 4
Find: a^3 + (1/a)^3
Solution: a + 1/a = -4
Take cube on both sides
(a + 1/a)^3 = (-4)^3
a^3 + (1/a)^3 + 3*a* 1/a (a+1/a) = -64
a^3 + (1/a)^3 + 3(a+1/a) = -64
a^3 + (1/a)^3 – 12 = -64
a^3 + (1/a)^3 = – 64 + 12
a^3 + (1/a)^3 = -52
Find: a^3 + (1/a)^3
Solution: a + 1/a = -4
Take cube on both sides
(a + 1/a)^3 = (-4)^3
a^3 + (1/a)^3 + 3*a* 1/a (a+1/a) = -64
a^3 + (1/a)^3 + 3(a+1/a) = -64
a^3 + (1/a)^3 – 12 = -64
a^3 + (1/a)^3 = – 64 + 12
a^3 + (1/a)^3 = -52
Correct Answer: A
Question 3 Explanation:
Given : a + 1/a = – 4
Find: a^3 + (1/a)^3
Solution: a + 1/a = -4
Take cube on both sides
(a + 1/a)^3 = (-4)^3
a^3 + (1/a)^3 + 3*a* 1/a (a+1/a) = -64
a^3 + (1/a)^3 + 3(a+1/a) = -64
a^3 + (1/a)^3 – 12 = -64
a^3 + (1/a)^3 = – 64 + 12
a^3 + (1/a)^3 = -52
Find: a^3 + (1/a)^3
Solution: a + 1/a = -4
Take cube on both sides
(a + 1/a)^3 = (-4)^3
a^3 + (1/a)^3 + 3*a* 1/a (a+1/a) = -64
a^3 + (1/a)^3 + 3(a+1/a) = -64
a^3 + (1/a)^3 – 12 = -64
a^3 + (1/a)^3 = – 64 + 12
a^3 + (1/a)^3 = -52
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