Predicting $x_t$ knowing something about $\Delta^2 x_t$ I have this exercise question, what is your prediction of $x_{10}$ knowing that $\Delta^2 x_t = \epsilon_t $, knowing $x_9 = 1.56$ and $x_8 = 1.64$.
I take this to mean that
$$
x_t = x_{t-2} +\epsilon_t
$$
where I assume the error term is iid with mean zero.
I would then answer that my best prediction is simply $x_8$ but the answer key gives a different answer of
$$
x_{10} = 2x_8 - x_9 = 1.72. 
$$

The first difference operator $\Delta$ is the difference between the identity $\mathbb I$ and the backshift operator $B$ given by
$$(Bx)_t = x_{t-1};$$
that is,
$$(\Delta x)_t = ((\mathbb I - B)x)_t = (\mathbb{I} x)_t - (Bx)_t = x_t - x_{t-1}.$$
Notice that iterations of $B$ shift the series even more:
$$(B^n x)_t = B(B^{n-1}x)_t = \cdots = x_{t-n}.$$
In particular, you have interpreted $\Delta^2$ as $B^2,$ which explains why your solution differs from the answer key.
The second difference operator $\Delta^2 = \Delta\circ\Delta$ results from two applications of $\Delta,$ whence
$$\left(\Delta^2x\right)_t = \left(\mathbb I - B)^2 x\right)_t = \left((\mathbb I - 2B + B^2)x\right)_t = x_t - 2x_{t-1} + x_{t-2}.$$
(Refer to Wikipedia for instance.)
We see that $\Delta^2$ differs from $B^2.$  Apply this result to $t=10$ to give
$$\epsilon_{10} = (\Delta^2 x)_{10} = x_{10} - 2x_9 + x_8 = x_{10} - 2(1.56) + 1.64.$$
Solving this equation yields a formula that differs from the solution in the question,
$$x_{10} = \epsilon_{10} + 2(1.56) - 1.64 = 1.48 + \epsilon_{10}.$$
I suspect you meant to stipulate that $x_9 = 1.64$ and $x_8=1.56$ rather than the other way around, so let's continue with that re-interpretation, which gives
$$x_{10} = 1.72 + \epsilon_{10}.$$
The best prediction is the conditional expectation
$$\hat x_{10} = E[\epsilon_{10} + 2(x_9) - x_8\mid x_8, x_9] = E[\epsilon_{10}] + 1.72 = 1.72$$
(because you have stated that the $\epsilon_t$ are independent and have expectations of zero).

For a fuller account of working with operators like $B$ and $\Delta$ see https://stats.stackexchange.com/a/568744/919.
