Showing a vectorspace equals a span of polynomials? I'm not sure on the titleing of this question, mostly because I don't really understand the question this task is asking me. Can someone point me in the right direction? 
Let $H_{2,3}\subset P_{2,3}$ be the vectorspace spanned by the polynomial on the form: $(a\cdot x_1 + b\cdot x_2)^3$ for $a,b\in \mathbb{R}$. Show that $H_{2,3} = P_{2,3}$.
Edit; it doesn't say what $P_{2,3}$ is, but guessing by the(as far as I can tell) unrelated previous task, it is denoting a vectorspace of homogenous polynomials of degree $3$ on $\mathbb{R}^2$

$H_{2,3}$ is the subspace of $P_{2,3}$ spanned by the polynomials of the form $(a x_1 + b x_2)^3$, with $a, b \in \mathbb R$. First note that $(a x_1 + b x_2)^3$ is indeed a homogeneous polynomial of degree 3, so $H_{2,3}$ is indeed a subset of $P_{2,3}$. Now, by definition,
$$H_{2,3} = \{ \sum_{i=1}^n (a_i x_1 + b_ix_2)^3 | a_1, b_1, \dots, a_n, b_n \in {\mathbb R}\}$$
and your task is to show that this is all of $P_{2,3}$.
The easiest way to go about this is, I think, to note that $P_{2,3}$ is spanned by $x_1^3, x_1^2 x_2, x_1 x_2^2, x_2^3$. Therefore it is enough to explicitly write these four polynomials as a linear combination of the generators of $H_{2,3}$. Now $x_1^3$ and $x_2^3$ are generators themselves (with $a = 1, b = 0$ and the other way around). By symmetry, this just leaves $x_1^2 x_2$.
