To find the intersection points, set the equations equal to each other:

\(x^{\frac{2}{3}} - 1 = x^{\frac{1}{3}} + 1\).

Rearrange to form:

\(x^{\frac{2}{3}} - x^{\frac{1}{3}} - 2 = 0\).

Let \(u = x^{\frac{1}{3}}\), then \(u^2 = x^{\frac{2}{3}}\).

The equation becomes:

\(u^2 - u - 2 = 0\).

Factor the quadratic:

\((u - 2)(u + 1) = 0\).

Thus, \(u = 2\) or \(u = -1\).

Convert back to \(x\):

For \(u = 2\), \(x = 2^3 = 8\).

For \(u = -1\), \(x = (-1)^3 = -1\).

Find \(y\) for each \(x\):

For \(x = 8\), \(y = 8^{\frac{1}{3}} + 1 = 2 + 1 = 3\).

For \(x = -1\), \(y = (-1)^{\frac{1}{3}} + 1 = -1 + 1 = 0\).

Thus, the points of intersection are \((8, 3)\) and \((-1, 0)\).