Evaluate a cube-root symmetric sum: If α and β are roots of 8x^2 − 3x + 27 = 0, find (α^2/β)^{1/3} + (β^2/α)^{1/3}.

Difficulty: Hard

Correct Answer: 1/4

Explanation:


Introduction / Context:
This is a nonstandard but elegant manipulation with roots of a quadratic. Define A = (α^2/β)^{1/3} and B = (β^2/α)^{1/3}. We are to find S = A + B. Using identities for A^3 + B^3 and AB, we can form a cubic in S and then locate the rational value among the options that satisfies it.


Given Data / Assumptions:

  • α + β = 3/8 (from −b/a)
  • αβ = 27/8 (from c/a)
  • A^3 = α^2/β, B^3 = β^2/α
  • AB = (αβ)^{1/3}


Concept / Approach:
We use (A + B)^3 = A^3 + B^3 + 3AB(A + B). Compute A^3 + B^3 in terms of α, β using the identity α^3 + β^3 = (α + β)^3 − 3αβ(α + β), then divide by αβ. Compute AB directly as (αβ)^{1/3}. Substitute into the cubic for S and solve by testing the given rational choices.


Step-by-Step Solution:

Let S1 = α + β = 3/8, P = αβ = 27/8α^3 + β^3 = S1^3 − 3PS1So A^3 + B^3 = (α^3 + β^3)/P = (S1^3 − 3PS1)/PCompute: S1^3 = 27/512; 3PS1 = 3*(27/8)*(3/8) = 243/64Hence A^3 + B^3 = (27/512 − 1944/512)/(27/8) = (−1917/512)*(8/27) = −71/64AB = P^{1/3} = (27/8)^{1/3} = 3/2Let S = A + B. Then S^3 = (−71/64) + 3*(3/2)*S = (−71/64) + (9/2)SMultiply by 64: 64S^3 − 288S + 71 = 0Test options: S = 1/4 satisfies 64*(1/64) − 288*(1/4) + 71 = 1 − 72 + 71 = 0


Verification / Alternative check:
None of the other options satisfy the cubic. Since the cubic has S = 1/4 as a root and other options are far off, 1/4 is the correct value.


Why Other Options Are Wrong:

  • 1/3, 7/2, 4, 3/4: Substitution into 64S^3 − 288S + 71 fails to give zero.


Common Pitfalls:
Arithmetic slips when converting to a common denominator or when dividing by P. Keep fractions exact; avoid decimals to prevent rounding errors.


Final Answer:
1/4

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