Question: I am currently trying to show that $$\{x^3+x^2,x^3+x\}$$ form a Basis for the subspace $$ W=\{p\in P\_3(\mathbb R):p(-1)=p(0)=0\}$$
and I can't pin down my mistake and would really appreciate some help. My 'solution' is as follows:

let $\ p \in W $: I need to show that $\forall\lambda\_1,\lambda\_2\in \mathbb R$:
$$ p=\lambda\_1(x^3+x^2)+\lambda\_2(x^3+x) $$
As $p\in W\ \exists a,b,c,d$ s.t.
$ p = ax^3+bx^2+cx+d$

$ p(0)=0$ s.t. $a0^3+b0^2+c0+d=0 \implies d=0$
therefore: $$
p(x) = ax^3+bx^2+cx$$
$p(-1) =0$ s.t. $a(-1)^3+b(-1)^2+c(-1) =0 \implies a =b-c$ therefore:
$$p(x)=(b-c)x^3+bx^2+cx\\p(x)=bx^3-cx^3+bx^2+cx\\p(x)=b(x^3+x^2)-c(x^3-x)
$$
which can't be since $x^3+x\ and\ x^3-x$ are linearly independent.
I've done this exercise in many different ways but there is always some little mistake where the sign of the second element does not match or where the degree of the second element does not match. I'm beginning to think there might be a typo in the exercise sheet. If anyone can confirm that, or confirm that I'm doing something wrong I would greatly appreciate it.

Answer: 
You can't prove that, since it is false. Note that $x^3+x\notin W$, since $(-1)^3+(-1)\neq0$.
I'll assume you want to show that $\{x^3+x^2,x^{\color{red}{2}}+x\}$ is a basis.

First, the two polynomials belong to $W$.

Now suppose $p(x)\in W$; write $p(x)=a+bx+cx^2+dx^3$; then the condition $p(x)\in W$ translates into $a=0$ and $a-b+c-d=0$, that is, $d=c-b$.

In particular, $p(x)=bx+cx^2+(c-b)x^3=b(x-x^3)+c(x^2+x^3)$ and therefore $W$ has dimension $2$.

Since $\{x^3+x^2,x^2+x\}$ is linearly independent (proof?), it is a basis.

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The typo in the book might be $\{x^3+x^2,x^3\color{red}{-}x\}$. The same reasoning as above applies.