The Coefficient of x⁷ in the Expansion of (1 − x − x² + x³)⁶ Is

AIEEE 2011 — Previous Year Question.

Problem

The coefficient of $x^7$ in the expansion of $(1-x-x^2+x^3)^6$ is:

Key insight. A four-term expression raised to the 6th power looks unwieldy to expand directly, but $1-x-x^2+x^3$ factors neatly into $(1-x)(1-x^2)$ — turning the whole problem into finding one coefficient in a product of two ordinary binomial expansions.

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Approach

Rather than expanding a four-term expression to the 6th power directly, factor $1-x-x^2+x^3$ by grouping — this reduces it to a product of two binomials, each of which is easy to raise to the 6th power using the standard binomial expansion. Then pick out every pair of terms — one from each expansion — whose powers of $x$ add up to 7, and sum their coefficients.

Solution

Step 1 — Factor the base by grouping

Group the four terms in pairs:

$$1-x-x^2+x^3 = (1-x)-x^2(1-x) = (1-x)(1-x^2)$$

Step 2 — Raise the factored form to the 6th power

$$(1-x-x^2+x^3)^6 = \left[(1-x)(1-x^2)\right]^6 = (1-x)^6(1-x^2)^6$$

This is now a product of two ordinary binomial expansions, each expandable using $\binom{6}{k}$.

Step 3 — Expand both factors

$$(1-x)^6 = \sum_{k=0}^{6}\binom6k(-1)^kx^k$$

$$(1-x^2)^6 = \sum_{j=0}^{6}\binom6j(-1)^jx^{2j}$$

Step 4 — Find every combination giving x⁷

A term $x^k$ from the first factor combines with a term $x^{2j}$ from the second to give $x^{k+2j}$. Since $k$ ranges over $0$ to $6$ and $2j$ is even, $k+2j=7$ forces $k$ to be odd — so the only possibilities with $k\le6$ are $k=1,3,5$:

$$k=1,\ j=3: \quad k+2j = 1+6 = 7$$ $$k=3,\ j=2: \quad k+2j = 3+4 = 7$$ $$k=5,\ j=1: \quad k+2j = 5+2 = 7$$

Step 5 — Compute each contributing product

For $k=1,\,j=3$: coefficient from $(1-x)^6$ is $\binom61(-1)^1=-6$; coefficient from $(1-x^2)^6$ is $\binom63(-1)^3=-20$.

$$(-6)\times(-20) = 120$$

For $k=3,\,j=2$: coefficient from $(1-x)^6$ is $\binom63(-1)^3=-20$; coefficient from $(1-x^2)^6$ is $\binom62(-1)^2=15$.

$$(-20)\times15 = -300$$

For $k=5,\,j=1$: coefficient from $(1-x)^6$ is $\binom65(-1)^5=-6$; coefficient from $(1-x^2)^6$ is $\binom61(-1)^1=-6$.

$$(-6)\times(-6) = 36$$

Step 6 — Add all three contributions

$$\text{Coefficient of }x^7 = 120-300+36 = -144$$

Answer

Common mistakes

• Trying to expand $(1-x-x^2+x^3)^6$ using the multinomial theorem directly. This works in principle but involves far more terms to track than the factored approach — the factoring step at the start is what makes the problem tractable by hand.
• Missing a valid (k,j) combination. It’s easy to check only $k=1$ and $k=3$ and overlook $k=5$ (paired with $j=1$), since $k=5$ is close to the upper limit of 6 and easy to skip past.
• Sign errors in $\binom6k(-1)^k$. With alternating signs on both factors, a single dropped negative sign in any of the three products changes the final answer’s sign or magnitude — it helps to compute the sign and the magnitude of each $\binom6k$ term separately before multiplying.

Practise next

• Find the coefficient of $x^8$ in the expansion of $(1-x-x^2+x^3)^6$, using the same factored form $(1-x)^6(1-x^2)^6$.