Question

Find the coefficient of \(x^4\) in \((x-1)^3(x-2)^3\).

• A. 33
• B. $-3$
• C. 30
• D. 21

Hint:

Two fast routes: (1) Pair terms so total exponents sum to the target degree; (2) First compress repeated factors (e.g., \((x-1)^3(x-2)^3=\bigl[(x-1)(x-2)\bigr]^3\)) and apply the multinomial to count only the degree-producing patterns.

Answer 1

The Correct Option is A

Method 1 (Targeted term pairing with Binomial Theorem).
Write \[ (x-1)^3=\sum_{k=0}^{3}\binom{3}{k}x^{3-k}(-1)^k, (x-2)^3=\sum_{j=0}^{3}\binom{3}{j}x^{3-j}(-2)^j. \] When multiplying, a general product term contributes degree \[ x^{(3-k)+(3-j)} = x^{6-(k+j)}. \] We need \(6-(k+j)=4\Rightarrow k+j=2\). The valid pairs \((k,j)\) are \((0,2),(1,1),(2,0)\). \begin{align*} (k,j)=(0,2):& \binom{3}{0}(-1)^0\cdot \binom{3}{2}(-2)^2 =1\cdot 3\cdot 4=12,
(k,j)=(1,1):& \binom{3}{1}(-1)^1\cdot \binom{3}{1}(-2)^1 =(-3)\cdot(-6)=18,
(k,j)=(2,0):& \binom{3}{2}(-1)^2\cdot \binom{3}{0}(-2)^0 =3\cdot 1=3. \end{align*} Sum of contributions \(\Rightarrow 12+18+3=33\). Hence the coefficient of \(x^4\) is \(\boxed{33}\). Method 2 (Compress first, then use multinomial).
Note \[ (x-1)^3(x-2)^3=\bigl[(x-1)(x-2)\bigr]^3=(x^2-3x+2)^3. \] Let \(a=x^2\), \(b=-3x\), \(c=2\). Expand \((a+b+c)^3\). To get \(x^4\), we need total degree \(4\). The only contributing patterns are: \begin{itemize} \item \(a^2c\): two \(x^2\) and one constant \(\Rightarrow x^4\). Multinomial count: \(\binom{3}{2,0,1}=3\). Contribution: \(3\cdot (x^2)^2\cdot 2=6x^4\). \item \(ab^2\): one \(x^2\) and two \((-3x)\) \(\Rightarrow x^2\cdot 9x^2=9x^4\). Count: \(\binom{3}{1,2,0}=3\). Contribution: \(3\cdot 9x^4=27x^4\). \end{itemize} Total \(x^4\)-coefficient \(=6+27=33\) \(\Rightarrow\boxed{33}\). Why no other patterns?
\(a^3\) gives \(x^6\); \(b^3\) gives \(x^3\); \(c^3\) gives degree \(0\); \(a^2b\) gives \(x^5\); \(abc\) gives \(x^3\); \(ac^2\) gives \(x^2\); \(b^2c\) gives \(x^2\). None equals degree \(4\). Final Answer:
\[ \boxed{33} \]