If $$(x + y)^{\frac{1}{3}} + (y + z)^{\frac{1}{3}} = -(z + x)^{\frac{1}{3}}$$, then $$(x^3 + y^3 + z^3)$$ can be expressed as:

A

$$\frac{1}{8} xyz$$

B

$$(x + y)(y + z)(z + x)$$

C

$$\frac{3}{8}(x + y)(y + z)(z + x)$$

D

$$3xyz$$