Check if polynomials vectors are linearly indpendent I need to know if $ {1+x,1-x,1-x^2} $ are linearly independent 
i know that i need to check it like
$$ a(1+x)+b(1-x)+c(1-x^2)=0 $$
In my math book
they show the solution
as $$ (a+b+c)+(a-b)x-cx^2=0 $$
but i didnt understand
why i can do this
$$ a+b+c =0 \\
a-b=0 \\
-c=0 $$
isnt it say by doing this that i check different polynomials? $${1,x,x^2}$$
These are different polynomials than those shown in the question, how can I actually do this? i mean how is this answer my quistion?
 A: Notice that the polynomials $p(x)=a_{0}+a_{1}x+a_{2}x^{2}$ and $q(x)=b_{0}+b_{1}x+b_{2}x^{2}$ they are equals if and only if the coefficients are equals, that is,
$$p(x)=q(x)\iff (a_0=b_0) \wedge (a_1=b_1)\wedge (a_2=b_2)$$
In your case in the rhs you have $0+0x+0x^{2}$.
A: When you it goes from $$a(1+x)+b(1-x)+c(1-x^2)=0$$ to $$(a+b+c)+(a-b)x-cx^2=0$$ it's using the distributive property. So for example, $a(1+x)=a+ax$. So using this the first equation becomes $$a+ax+b-bx+c-cx^2=0.$$ Now we can add terms in any order so we rearrange to $$(a+b+c)+ax-bx-cx^2=0.$$ Notice I grouped the terms without an $x$, the terms with $x$ and the term with $x^2$ is on its own. Now we factor an $x$ in the middle to get $$(a+b+c)+(a-b)x-cx^2=0.$$ Now we're not checking independence of $1,x,x^2$. All we've shown is that if $$a(1+x)+b(1-x)+c(1-x^2)=0$$ then $(a+b+c)+(a-b)x-cx^2=0$.
Now we will use the crucial fact that $1,x,x^2$ are indeed linearly independent. This is where we get the system of equations in your post and solving that gives us $a=b=c=0$.
