Find the coefficient of $x^3$ in the expansion of $(1+x+x^2)^3$ using Binomial theorem.

Question

Find the coefficient of $x^3$ in the expansion of $(1+x+x^2)^3$ using Binomial theorem.

Answer 1

Every term in the expansion is the result of the product of the three terms $1,x,x^2$ with some exponents and we can achieve $x^3$ as

$$1^0\,x^3\,(x^2)^0\text{ or }1^0\,x^1\,(x^2)^1.$$

These terms appear with counts given by the multinomial coefficient, and

$$n=\frac{3!}{0!\,3\,!0!}+\frac{3!}{0!\,1!\,1!}=7.$$

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For information,

$$(1+x+x^2)^3=x^6+3x^5+6x^4+7x^3+6x^2+3x+1.$$

Answer 2

Hint:

For $x\ne1,$ $$(1+x+x^2)^3=\left(\dfrac{1-x^3}{1-x}\right)^3=(1-x^3)^3(1-x)^{-3}$$

Now use Binomial series to find the coefficient of $x^3$ to be

$1\cdot$the coefficient of $x^3$ in $(1-x)^{-3}-3\cdot$the coefficient of $x^{3-3}$ in $(1-x)^{-3}$

$$=1\cdot\dfrac{(-3)(-3-1)(-3-2)}{3!}(-1)^3-3\cdot1=\dfrac{3\cdot4\cdot5}{1\cdot2\cdot3}-3$$

Answer 3

Espress it as a product of three brackets: $$(1+x+x^2)^3=(\underbrace{1+x+x^2}_{{x^i}})(\underbrace{1+x+x^2}_{{x^j}})(\underbrace{1+x+x^2}_{{x^k}})$$ Note that $i+j+k=3$. The possible triples $(i,j,k)$ are: $$(0,1,2),(0,2,1),(1,0,2),(1,2,0),(2,0,1),(2,1,0),(1,1,1).$$ Hence, the coefficient of $x^3$ is $7$.