If $$z = x - iy$$ and $$z^{1/3} = p + iq$$, then $$\frac{\left(\frac{x}{p} + \frac{y}{q}\right)}{(p^2 + q^2)}$$ is equal to

Given the relation between the complex numbers:

$$ z^{1/3} = p + iq $$

Cube both sides of the equation to express z in terms of p and q:

$$ z = (p + iq)^3 $$

Expand the right side using the binomial cube expansion formula:

$$ z = p^3 + 3p^2(iq) + 3p(iq)^2 + (iq)^3 $$

Substitute the values for the powers of the imaginary unit where $$ i^2 = -1 $$ and $$ i^3 = -i $$:

$$ z = p^3 + 3ip^2q - 3pq^2 - iq^3 $$

Group the real terms together and the imaginary terms together:

$$ z = (p^3 - 3pq^2) + i(3p^2q - q^3) $$

The problem defines z as:

$$ z = x - iy $$

Equate the real and imaginary parts from both expressions of z:

$$ x = p^3 - 3pq^2 $$

$$ -y = 3p^2q - q^3 $$

Multiply the second equation by -1 to isolate y:

$$ y = q^3 - 3p^2q $$

Divide the equation for x by p to find the first required ratio:

$$ \frac{x}{p} = p^2 - 3q^2 $$

Divide the equation for y by q to find the second required ratio:

$$ \frac{y}{q} = q^2 - 3p^2 $$

Add the two resulting expressions together:

$$ \frac{x}{p} + \frac{y}{q} = (p^2 - 3q^2) + (q^2 - 3p^2) $$

Combine the like terms on the right side:

$$ \frac{x}{p} + \frac{y}{q} = -2p^2 - 2q^2 $$

Factor out the common multiplier -2:

$$ \frac{x}{p} + \frac{y}{q} = -2(p^2 + q^2) $$

Divide both sides by the quantity in parentheses to match the required fraction form:

$$ \frac{\left(\frac{x}{p} + \frac{y}{q}\right)}{(p^2 + q^2)} = -2 $$

Final Answer:

The value of the expression is -2.