Answer
$-16y^{3}+3y^{2}+y+6$
Work Step by Step
The difference between $(-6+y^{2}+5y^{3})$ and $(-12-y+13y^{3})$,
after writing each polynomial in standard form
( in order of powers of y, highest first)
is
$ \begin{array}{rrrrrr}
& 5y^{3} & +y^{2} & & -6\ \ \\
-( & 13y^{3} & & -y & -12)\\
\hline & & & & \\
= & -8y^{3} & +y^{2} & +y & +6
\end{array}$
From this, we subtract $(-2y^{2}+8y^{3})$,
(writing in standard form)
$ \begin{array}{rrrrrrr}
& -8y^{3} & +y^{2} & +y & +6\\
-( & 8y^{3} & -2y^{2} & & )\\
\hline & & & & \\
= & -16y^{3} & +3y^{2} & +y & +6\\
& & & &
\end{array}$
