1. **State the problem:** Evaluate the expression $$(1 - j)^3$$ where $j$ is the imaginary unit with the property $j^2 = -1$. 2. **Recall the binomial expansion formula:** For any complex numbers $a$ and $b$, $$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$$ 3. **Apply the formula:** Let $a = 1$ and $b = -j$, then $$ (1 - j)^3 = 1^3 + 3(1)^2(-j) + 3(1)(-j)^2 + (-j)^3 $$ 4. **Calculate each term:** - $1^3 = 1$ - $3(1)^2(-j) = 3 imes 1 imes (-j) = -3j$ - $3(1)(-j)^2 = 3 imes 1 imes (-j)^2$ Since $(-j)^2 = (-1)^2 imes j^2 = 1 imes (-1) = -1$, so $$3 imes 1 imes (-1) = -3$$ - $(-j)^3 = (-j) imes (-j)^2 = (-j) imes (-1) = j$ 5. **Sum all terms:** $$1 - 3j - 3 + j = (1 - 3) + (-3j + j) = -2 - 2j$$ 6. **Final answer:** $$(1 - j)^3 = -2 - 2j$$ 7. **Check options:** None of the options exactly match $-2 - 2j$, so the correct evaluation is $-2 - 2j$ which is not listed among the options.