Solution:

Let $a_n = a_1 + (n-1)d$ for some $a_1$ and common difference $d$.

First, $a_1 + a_2 + \cdots + a_{10} = 20$.

The sum of the first $10$ terms of an arithmetic sequence is:
$$S_{10} = \frac{10}{2}(a_1 + a_{10}) = 5(a_1 + a_{10})$$
But $a_{10} = a_1 + 9d$, so:
$$S_{10} = 5(a_1 + a_1 + 9d) = 5(2a_1 + 9d) = 10a_1 + 45d$$
Set equal to $20$:
$$10a_1 + 45d = 20$$

Second, $a_1 + a_4 + a_9 + a_{16} + \cdots + a_{100}$.

These are the terms $a_{n^2}$ for $n = 1$ to $10$ ($n^2$ from $1$ to $100$).

$a_{n^2} = a_1 + (n^2 - 1)d$

So:
$$\sum_{n=1}^{10} a_{n^2} = \sum_{n=1}^{10} [a_1 + (n^2 - 1)d] = 10a_1 + d \sum_{n=1}^{10} (n^2 - 1)$$
$$= 10a_1 + d \left( \sum_{n=1}^{10} n^2 - 10 \right)$$
The sum $\sum_{n=1}^{10} n^2 = \frac{10 \cdot 11 \cdot 21}{6} = 385$

So:
$$10a_1 + d(385 - 10) = 10a_1 + 375d$$
Set equal to $18$:
$$10a_1 + 375d = 18$$

Now subtract the first equation from the second:
$$(10a_1 + 375d) - (10a_1 + 45d) = 18 - 20$$
$$330d = -2$$
$$d = -\frac{2}{330} = -\frac{1}{165}$$

Now plug $d$ back into the first equation:
$$10a_1 + 45d = 20$$
$$10a_1 + 45 \left(-\frac{1}{165}\right) = 20$$
$$10a_1 - \frac{45}{165} = 20$$
$$10a_1 - \frac{9}{33} = 20$$
$$10a_1 = 20 + \frac{9}{33} = \frac{660}{33} + \frac{9}{33} = \frac{669}{33}$$
$$a_1 = \frac{669}{330}$$

Now, compute $a_1 + a_8 + a_{27} + a_{64} + \cdots + a_{1000}$.

These are $a_{n^3}$ for $n = 1$ to $10$ ($n^3$ from $1$ to $1000$).

$a_{n^3} = a_1 + (n^3 - 1)d$

So:
$$\sum_{n=1}^{10} a_{n^3} = 10a_1 + d \sum_{n=1}^{10} (n^3 - 1)$$
$$= 10a_1 + d \left( \sum_{n=1}^{10} n^3 - 10 \right)$$
The sum $\sum_{n=1}^{10} n^3 = \left( \frac{10 \cdot 11}{2} \right)^2 = 55^2 = 3025$

So:
$$10a_1 + d(3025 - 10) = 10a_1 + 3015d$$
Now substitute $a_1 = \frac{669}{330}$ and $d = -\frac{1}{165}$:

$10a_1 = \frac{669}{33}$

$3015d = 3015 \times \left(-\frac{1}{165}\right) = -\frac{3015}{165} = -\frac{3015}{165}$

Simplify $\frac{3015}{165}$:
$3015 \div 165 = 18.27$ but let's do it exactly:
$3015 \div 165 = \frac{3015}{165} = \frac{3015 \div 15}{165 \div 15} = \frac{201}{11}$

So $3015d = -\frac{201}{11}$

Now add:
$$\sum_{n=1}^{10} a_{n^3} = \frac{669}{33} - \frac{201}{11}$$
Find common denominator:
$33$ and $11$; $33$ is common.
$\frac{201}{11} = \frac{201 \times 3}{33} = \frac{603}{33}$

So:
$$\frac{669}{33} - \frac{603}{33} = \frac{66}{33} = 2$$

Answer: $a_1 + a_8 + a_{27} + a_{64} + \cdots + a_{1000} = 2$