i am looking at an answer to an exercise who asks to find a linear span for, and I don't really understand the solution
$$p(x) = ax^3 + bx ^2 + cx + d$$
and this is the solution i see
$$ p(x) \in M \leftrightarrow p(x) = p(x-1)$$ $$ \leftrightarrow ax^3 + bx^2 + cx +d = a(x-1)^3 + b(x-1)^2 + c(x-1) + d $$ $$\leftrightarrow ax^3 + bx^2 + cx + d = ax^3 + (b-3a)x^2 + (3a-2b+c)x+b-a-c+d$$ $$\leftrightarrow b-3a = b, 3a-2b+c = c, b-a-c+d = d $$ $$ \leftrightarrow a=b=c=0 $$ $$p(x) = d$$ $$Sp\{1\} = \{p(x) = d | d \in \mathbb{R}\} $$ now the thing that I don't really understand start from the third line how did they reach this equation? $$ax^3 + (b-3a)x^2 + (3a-2b+c)x+b-a-c+d$$ and how did they find the span from it?
They are using
$a(x-1)^3+b(x-1)^2+c(x-1)+d=a(x^3-3x^2+3x-1)+b(x^2-2x+1)+cx-c+d$
$\hspace{2.5 in}=ax^3+(b-3a)x^2+(3a-2b+c)x+b-a-c+d$,
so setting the coefficients of the same powers of x equal to each other gives
$\;\;\;b-3a=b, \;\;c=3a-2b+c, \;\;d=b-a-c+d$.
Then $a=0$ from the 1st equation, so $b=0$ from the 2nd equation, and then $c=0$ from the 3rd equation.
Therefore $p(x)=ax^3+bx^2+cx+d=0x^3+0x^2+0x+d=d=d(1)$,
so the vector $\{1\}$ spans the subspace M.