Surd ratios and symmetric powers Let x = (√3 + 1)/(√3 − 1) and y = (√3 − 1)/(√3 + 1). Evaluate x^2/y + y^2/x.

Introduction / Context:
Expressions featuring x and its reciprocal often reduce via symmetry. Here, y is the reciprocal of x, enabling a simplification from mixed fractions to a sum of cubes of x and 1/x.

Given Data / Assumptions:

• x = (√3 + 1)/(√3 − 1), y = (√3 − 1)/(√3 + 1).
• We need S = x^2/y + y^2/x.

Concept / Approach:
Notice y = 1/x. Then S = x^2 * x + (1/x^2) * (1/x) = x^3 + 1/x^3. Next compute x in simplest form, then use the identity x^3 + 1/x^3 = (x + 1/x)^3 − 3(x + 1/x).

Step-by-Step Solution:
Simplify x: (√3 + 1)/(√3 − 1) = ((√3 + 1)^2)/(3 − 1) = (3 + 1 + 2√3)/2 = 2 + √3.Thus 1/x = 2 − √3 (since (2 + √3)(2 − √3) = 1).Then x + 1/x = (2 + √3) + (2 − √3) = 4.Compute x^3 + 1/x^3 = (x + 1/x)^3 − 3(x + 1/x) = 4^3 − 3*4 = 64 − 12 = 52.

Verification / Alternative check:
Directly cube x and 1/x numerically using √3 ≈ 1.732; sums confirm the exact value 52.

Why Other Options Are Wrong:
76 and 64 come from squaring 4 or forgetting the subtraction term; 4 corresponds to x + 1/x, not the cube sum; 28 is a rough, incorrect estimate.

Common Pitfalls:
Not noticing y = 1/x; failing to rationalize x first, which hides the elegant cancellation.

Final Answer:
52