Binomial Expansion of $$(1-x)^n$$

First factor the expression:

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

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

Therefore,

$$(1-x-x^2+x^3)^6=(1-x)^6(1-x^2)^6$$

Now,

$$(1-x)^6=\sum_{r=0}^{6}(-1)^r\binom{6}{r}x^r$$

and

$$(1-x^2)^6=\sum_{k=0}^{6}(-1)^k\binom{6}{k}x^{2k}$$

To obtain the coefficient of

$$x^7$$

we require

$$r+2k=7$$

The possible pairs are

$$(k,r)=(1,5),\ (2,3),\ (3,1)$$

For

$$(k,r)=(1,5)$$

the contribution is

$$(-1)^1\binom61\cdot(-1)^5\binom65$$

$$=(-6)(-6)$$

$$=36$$

For

$$(k,r)=(2,3)$$

the contribution is

$$(-1)^2\binom62\cdot(-1)^3\binom63$$

$$=(15)(-20)$$

$$=-300$$

For

$$(k,r)=(3,1)$$

the contribution is

$$(-1)^3\binom63\cdot(-1)^1\binom61$$

$$=(-20)(-6)$$

$$=120$$

Hence the required coefficient is

$$36-300+120$$

$$=-144$$

Final Answer :

$$-144$$