It's easy to show that the polynomial 1 is not in the given vector space (that is, it's easy to show that there are no real numbers $a,b$ such that $a(1+x+x^2)+b(x-x^2-x^3)$ is identically 1 --- can you do this?), and it's just as easy to show that $x^3$ is not in the space spanned by $1,1+x+x^2,x-x^2-x^3$ so you can take as your complement the space spanned by 1 and $x^3$ (in your notation, ${\bf R}(1)+{\bf R}(x^3)$).
A more systematic approach would be to form the matrix $\pmatrix{1&1&1&0\cr0&1&-1&-1\cr}$ whose rows come in what I hope is the obvious way from your given polynomials $1+x+x^2$ and $x-x^2-x^3$, and then find a basis for the nullspace of this matrix. Have you learned how to do that? You can bring the matrix to reduced row-echelon form, $\pmatrix{1&0&2&1\cr0&1&-1&-1\cr}$ and the read out a basis for the nullspace as $\{{(-2,1,1,0),(-1,1,0,1)\}}$ and then interpret these vectors as the polynomials $-2+x+x^2,-1+x+x^3$. Then ${\bf R}(-2+x+x^2)+{\bf R}(-1+x+x^3)$ is your complement.
This is not the same as the first answer I gave, but that's OK --- a given vector space has lots of complements.