Given a^2 + 1/a^2 = 17/4, determine the exact value of a^3 − 1/a^3. Assume real a for principal values.

Difficulty: Medium

Correct Answer: 63/8

Explanation:


Introduction / Context:
This problem explores classic symmetric identities with a and 1/a. Starting from a^2 + 1/a^2, you can recover both a + 1/a and a − 1/a and then compute higher-order expressions such as a^3 − 1/a^3 using factorization formulas.



Given Data / Assumptions:

  • a^2 + 1/a^2 = 17/4.
  • a is real and non-zero (so reciprocals exist).
  • We want a^3 − 1/a^3.


Concept / Approach:
Use (a ± 1/a)^2 relations: a^2 + 1/a^2 = (a + 1/a)^2 − 2 = (a − 1/a)^2 + 2. Then use a^3 − 1/a^3 = (a − 1/a)(a^2 + 1/a^2 + 1). Choose the principal positive root for a − 1/a to obtain the standard positive value.



Step-by-Step Solution:

(a + 1/a)^2 = 17/4 + 2 = 25/4 ⇒ a + 1/a = ±5/2.(a − 1/a)^2 = 17/4 − 2 = 9/4 ⇒ a − 1/a = ±3/2.Compute a^3 − 1/a^3 = (a − 1/a)(a^2 + 1/a^2 + 1) = (±3/2)(17/4 + 1) = (±3/2)(21/4) = ±63/8.Taking the principal positive value gives 63/8.


Verification / Alternative check:
If a − 1/a is taken negative, the magnitude remains 63/8 but the sign flips. Most exam conventions assume the positive branch unless specified otherwise.



Why Other Options Are Wrong:
75/16 and 95/8 come from misusing the identities or adding 1 incorrectly. “None of these” is incorrect because 63/8 is attainable directly.



Common Pitfalls:
Confusing a^3 − 1/a^3 with a^3 + 1/a^3, and misapplying the ± when taking square roots. Always plug back to confirm.



Final Answer:
63/8

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