We are given that $(x + iy)^3 = u + iv$.

Step 1: Expand the complex number First, we expand $(x + iy)^3$ using the binomial expansion as shown in Example 3 of the text:

$$(x + iy)^3 = x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3$$ $$= x^3 + 3ix^2y + 3x(-y^2) + (-i y^3)$$ $$= x^3 + 3ix^2y - 3xy^2 - iy^3$$

Grouping the real and imaginary parts, we get:

$$(x^3 - 3xy^2) + i(3x^2y - y^3)$$

Step 2: Equate real and imaginary parts Comparing the real and imaginary parts with $u + iv$ (similar to Example 1), we find:

$$u = x^3 - 3xy^2$$ $$v = 3x^2y - y^3$$

Step 3: Substitute values into the expression Now, we substitute these values of $u$ and $v$ into the left-hand side of the required expression:

$$\frac{u}{x} + \frac{v}{y} = \frac{x^3 - 3xy^2}{x} + \frac{3x^2y - y^3}{y}$$

Step 4: Simplify the expression Simplifying the fractions:

$$= (x^2 - 3y^2) + (3x^2 - y^2)$$ $$= x^2 + 3x^2 - 3y^2 - y^2$$ $$= 4x^2 - 4y^2$$ $$= 4(x^2 - y^2)$$

Thus, we have shown that $\frac{u}{x} + \frac{v}{y} = 4(x^2 - y^2)$.