Evaluate (1-2i)⁻³

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

$$(1-2i)^3 = 1^3 - 3(1)^2(2i) + 3(1)(2i)^2 - (2i)^3$$ $$= 1 - 6i + 3(4i^2) - 8i^3$$

Substitute $i^2 = -1$ and $i^3 = i^2 \cdot i = -i$:

$$= 1 - 6i + 3(4(-1)) - 8(-i)$$ $$= 1 - 6i - 12 + 8i$$

Combine the real and imaginary parts:

$$= (1-12) + (-6+8)i$$ $$= -11 + 2i$$

The conjugate of $(-11+2i)$ is $(-11-2i)$.

$$\frac{1}{-11+2i} \times \frac{-11-2i}{-11-2i}$$

The numerator is $1 \times (-11-2i) = -11-2i$.
The denominator is of the form $(a+bi)(a-bi) = a^2+b^2$. Here $a=-11$ and $b=2$.

$$(-11)^2 + (2)^2 = 121 + 4 = 125$$

So the expression becomes:

$$\frac{-11-2i}{125}$$

$$-\frac{11}{125} - \frac{2}{125}i$$

Comparing this with the given options, it matches option (a).

The final answer is $\text{(a) } \left(-\frac{11}{125} - \frac{2}{125}i\right)$.