34 (1276). For what values of $a, b, c$ and $d$ is the equality $5 x^{3}-32 x^{2}+75 x-71=a(x-2)^{3}+b(x-2)^{2}+c(x-2)+d$ an identity
Solution. We will expand the brackets on the right side of the equality and combine like terms:

$$
\begin{gathered}
a(x-2)^{3}+b(x-2)^{2}+c(x-2)+d= \\
=a x^{3}-6 a x^{2}+12 a x-8 a+b x^{2}-4 b x+4 b+c x-2 c+d= \\
=a x^{3}+(b-6 a) x^{2}+(12 a-4 b+c) x-8 a+4 b-2 c+d
\end{gathered}
$$

52

For the equality

$$
\begin{aligned}
& 5 x^{3}-32 x^{2}+75 x-71=a x^{3}+(b-6 a) x^{2}+(12 a-4 b+c) x-8 a+ \\
&+4 b-2 c+d
\end{aligned}
$$

to be an identity, it is necessary that the coefficients of the same powers coincide. Equating them, we obtain the system of equations:

$$
\left\{\begin{array}{l}
a=5 \\
b-6 a=-32 \\
12 a-4 b+c=75 \\
-8 a+4 b-2 c+d=-71
\end{array}\right.
$$

Solving it, we get that for $a=5, b=-2, c=7, d=-9$ the given equality will be an identity.

Answer: $a=5, b=-2, c=7, d=-9$.

3 remarks. It is also advisable to solve problems 1073,1074 from $[4]$.