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Problem 4 - Entrance Test

If x = a cos³θ and y = a sin³θ, then find the value of (x^(2/3) + y^(2/3)).

Correct: A

Given x = a cos³θ, divide by a: x/a = cos³θ. Taking the cube root on both sides: (x/a)^(1/3) = cos θ. Squaring both sides: x^(2/3)/a^(2/3) = cos²θ. Given y = a sin³θ, divide by a: y/a = sin³θ. Taking the cube root on both sides: (y/a)^(1/3) = sin θ. Squaring both sides: y^(2/3)/a^(2/3) = sin²θ. Now, add the two squared equations: x^(2/3)/a^(2/3) + y^(2/3)/a^(2/3) = cos²θ + sin²θ. Factor out 1/a^(2/3) on the left side: (x^(2/3) + y^(2/3))/a^(2/3) = cos²θ + sin²θ. Using the identity cos²θ + sin²θ = 1: (x^(2/3) + y^(2/3))/a^(2/3) = 1. Multiply by a^(2/3): x^(2/3) + y^(2/3) = a^(2/3).