If the curves $$x = y^4$$ and $$xy = k$$ cut at right angles, then $$(4k)^6$$ is equal to ______.

We need to find where $$x = y^4$$ and $$xy = k$$ intersect at right angles. Differentiating $$x = y^4$$ with respect to $$x$$: $$1 = 4y^3 \frac{dy}{dx}$$, so $$\frac{dy}{dx} = \frac{1}{4y^3}$$.

Differentiating $$xy = k$$: $$y + x\frac{dy}{dx} = 0$$, so $$\frac{dy}{dx} = -\frac{y}{x}$$.

For the curves to cut at right angles, the product of their slopes must be $$-1$$: $$\frac{1}{4y^3} \times \left(-\frac{y}{x}\right) = -1$$. This simplifies to $$\frac{1}{4y^2 x} = 1$$, giving $$4y^2 x = 1$$.

At the point of intersection, $$x = y^4$$. Substituting: $$4y^2 \cdot y^4 = 1$$, so $$4y^6 = 1$$, which gives $$y^6 = \frac{1}{4}$$.

Also, $$k = xy = y^4 \cdot y = y^5$$, so $$k^6 = y^{30} = (y^6)^5 = \left(\frac{1}{4}\right)^5 = \frac{1}{1024}$$.

Therefore, $$(4k)^6 = 4^6 \cdot k^6 = 4096 \times \frac{1}{1024} = 4$$.