Answer
$z=\frac{1}{17}(x^3+y^2)$
Work Step by Step
Step 1. Write the general form as $z=k(x^3+y^2)$ where $k$ is a constant.
Step 2. Use the given condition, $1=k(2^3+3^2)$, thus $k=\frac{1}{17}$
Step 3. We have $z=\frac{1}{17}(x^3+y^2)$
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