Let the area of the triangle with vertices $$A(1, \alpha)$$, $$B(\alpha, 0)$$ and $$C(0, \alpha)$$ be $$4$$ sq. units. If the points $$(\alpha, -\alpha)$$, $$(-\alpha, \alpha)$$ and $$(\alpha^2, \beta)$$ are collinear, then $$\beta$$ is equal to

The area of the triangle

$$ABC$$

is

$$\frac12\left|1(0-\alpha)+\alpha(\alpha-\alpha)+0(\alpha-0)\right|=4.$$

Thus,

$$\frac12|\,-\alpha\,|=4,$$

which gives

$$|\alpha|=8.$$

Hence

$$\alpha=\pm 8.$$

Now, the points

$$(\alpha,-\alpha)$$

and

$$(-\alpha,\alpha)$$

lie on the line

$$y=-x,$$

since

$$-\alpha=-\alpha$$

and

$$\alpha=-(-\alpha).$$

Therefore, for the three points to be collinear, the point

$$(\alpha^2,\beta)$$

must also lie on

$$y=-x.$$

Hence,

$$\beta=-\alpha^2.$$

Since

$$|\alpha|=8,$$

we have

$$\alpha^2=64.$$

Therefore,

$$\beta=-64.$$

Hence, $$\text{Answer}=-64$$.