Expand the denominator (1-i)³.

We use the binomial expansion formula $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$.
Here, $a=1$ and $b=i$.

$$(1-i)^3 = 1^3 - 3(1)^2(i) + 3(1)(i)^2 - i^3$$

Recall that $i^2 = -1$ and $i^3 = i^2 \cdot i = (-1)i = -i$.

$$(1-i)^3 = 1 - 3i + 3(-1) - (-i)$$ $$(1-i)^3 = 1 - 3i - 3 + i$$

Combine the real and imaginary parts:

$$(1-i)^3 = (1-3) + (-3i+i)$$ $$(1-i)^3 = -2 - 2i$$

The expression becomes:

$$\frac{4+3i}{-2-2i}$$

To rationalize, multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of $-2-2i$ is $-2+2i$.

$$\frac{4+3i}{-2-2i} \times \frac{-2+2i}{-2+2i}$$

$$(4+3i)(-2+2i) = 4(-2) + 4(2i) + 3i(-2) + 3i(2i)$$ $$= -8 + 8i - 6i + 6i^2$$

Substitute $i^2 = -1$:

$$= -8 + 2i + 6(-1)$$ $$= -8 + 2i - 6$$ $$= -14 + 2i$$

This is in the form $(a-b)(a+b) = a^2 - b^2$.

$$(-2-2i)(-2+2i) = (-2)^2 - (2i)^2$$ $$= 4 - (4i^2)$$

Substitute $i^2 = -1$:

$$= 4 - 4(-1)$$ $$= 4 + 4$$ $$= 8$$

$$\frac{-14}{8} + \frac{2i}{8}$$ $$-\frac{7}{4} + \frac{1}{4}i$$