Expanded by the binomial theorem in ascending powers of $x$ up to $x^3$, $\dfrac{1}{(4-x)^2}$ is:

1. Factor the 4 out: $\dfrac{1}{(4-x)^2}=\dfrac{1}{16}\left(1-\dfrac{x}{4}\right)^{-2}$. 2. Apply $(1+u)^n=1+nu+\dfrac{n(n-1)u^2}{2!}+\dfrac{n(n-1)(n-2)u^3}{3!}$ with $n=-2$ and $u=-\dfrac{x}{4}$. 3. The bracket becomes $1+\dfrac{x}{2}+\dfrac{3x^2}{16}+\dfrac{x^3}{16}$ — every sign is positive because $n$ is negative and $u$ is negative. 4. Multiply through by $\dfrac{1}{16}$. 5. This gives $\dfrac{1}{16}+\dfrac{x}{32}+\dfrac{3x^2}{256}+\dfrac{x^3}{256}$. The alternating-sign option forgets that the two negatives cancel. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/2, Question 6: Algebra_