Step 1: 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: $$ (1-i)^3 = (1-3) + (-3i+i) $$ $$ (1-i)^3 = -2 - 2i $$ Step 2: Substitute the expanded denominator back into the expression. The expression becomes: $$ \frac{4+3i}{-2-2i} $$ Step 3: Rationalize the denominator. To rationalize, multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of $-2-2i$ is $-2+2i$. $$ \frac{4+3i}{-2-2i} \times \frac{-2+2i}{-2+2i} $$ Step 4: Multiply the numerators. $$ (4+3i)(-2+2i) = 4(-2) + 4(2i) + 3i(-2) + 3i(2i) $$ $$ = -8 + 8i - 6i + 6i^2 $$ Substitute $i^2 = -1$: $$ = -8 + 2i + 6(-1) $$ $$ = -8 + 2i - 6 $$ $$ = -14 + 2i $$ Step 5: Multiply the denominators. 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 $i^2 = -1$: $$ = 4 - 4(-1) $$ $$ = 4 + 4 $$ $$ = 8 $$ Step 6: Combine the simplified numerator and denominator. $$ \frac{-14+2i}{8} $$ Step 7: Simplify the fraction by dividing both the real and imaginary parts by 8. $$ \frac{-14}{8} + \frac{2i}{8} $$ $$ -\frac{7}{4} + \frac{1}{4}i $$ The final answer is $\boxed{-\frac{7}{4} + \frac{1}{4}i}$.