35 (1277). Represent the polynomial $3 x^{3}+7 x^{2}+9 x+6$ as a polynomial $a y^{3}+b y^{2}+c y+d$, where $y=x+1$.
Solution. Method I. From the equality $y=x+1$, express $x$ and substitute the value $x=y-1$ into the given polynomial:

$$
\begin{gathered}
3 x^{3}+7 x^{2}+9 x+6=3(y-1)^{3}+7(y-1)^{2}+9(y-1)+6= \\
=3\left(y^{3}-3 y^{2}+3 y-1\right)+7\left(y^{2}-2 y+1\right)+9(y-1)+6= \\
=3 y^{3}-2 y^{2}+4 y+1 .
\end{gathered}
$$

We obtained $3 y^{3}-2 y^{2}+4 y+1=a y^{3}+b y^{2}+c y+d$, i.e., $a=3, b=-2$, $c=4, d=1$.

Method II. It is necessary to find such values of $a, b, c$ and $d$ so that the equality

$$
3 x^{3}+7 x^{2}+9 x+6=a(x+1)^{3}+b(x+1)^{2}+c(x+1)+d
$$

is an identity.

We solve using the method applied in solving problem 34:

$$
\begin{gathered}
a(x+1)^{3}+b(x+1)^{2}+c(x+1)+d=a x^{3}+3 a x^{2}+3 a x+a+b x^{2}+ \\
+2 b x+b+c x+c+d= \\
=a x^{3}+(3 a+b) x^{2}+(3 a+2 b+c) x+a+b+c+d
\end{gathered}
$$

We obtained

$$
\begin{gathered}
a x^{3}+(3 a+b) x^{2}+(3 a+2 b+c) x+a+b+c+d= \\
=3 x^{3}+7 x^{2}+9 x+6
\end{gathered}
$$

We have the system of equations:

$$
\left\{\begin{array}{l}
a=3 \\
3 a+b=7 \\
3 a+2 b+c=9 \\
a+b+c+d=6
\end{array}\right.
$$

from which $a=3, b=-2, c=4, d=1$.

Answer: $\quad 3 x^{3}+7 x^{2}+9 x+6=3(x+1)^{3}-2(x+1)^{2}+4(x+1)+1$.