Solve for a^3 from a reciprocal linear relation: If a + 1/a = 1 (a ≠ 0), determine the value of a^3.
Aptitude
Elementary Algebra
Difficulty: Medium
Choose an option
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A– 2
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B2
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C– 1
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D4
Answer
Correct Answer: – 1
Explanation
Introduction / Context:The relation a + 1/a = 1 leads to a quadratic equation in a. From that, powers of a can be reduced systematically. Here we are asked for a^3, which can be found without computing any decimals by using the quadratic to eliminate higher powers.
Given Data / Assumptions:
- a + 1/a = 1
- a ≠ 0
- Find a^3
Concept / Approach:Multiply the given equation by a to form a quadratic: a^2 − a + 1 = 0. Then express a^2 in terms of a and substitute into a^3 = a*a^2 to reduce a^3 to a linear expression in a, which then collapses to a constant using the same quadratic again.
Step-by-Step Solution:
From a + 1/a = 1 ⇒ multiply by a: a^2 + 1 = a ⇒ a^2 − a + 1 = 0.Hence a^2 = a − 1.Compute a^3 = a * a^2 = a(a − 1) = a^2 − a.Replace a^2 using a^2 − a + 1 = 0 ⇒ a^2 = a − 1.Thus a^3 = (a − 1) − a = −1.Verification / Alternative check:
The roots of a^2 − a + 1 = 0 are complex conjugates with magnitude 1; their cube equals −1, consistent with the algebraic reduction.Why Other Options Are Wrong:
- 2, 4, −2: None satisfy the derived identity; substituting back into a + 1/a = 1 fails.
Common Pitfalls:
- Mistaking a^2 − a + 1 = 0 for the cube roots of unity equation a^2 + a + 1 = 0.
- Arithmetic slips when computing a^3 = a(a − 1).
Final Answer:
– 1