Find $x$

Since the number is greater than 1000 and it has more than 3 digits it will be of the form 1000a + 100b + 10c + d.

b, c, and d should be between 0 and 9 inclusive but a can be greater than 10.

If we remove the last 3 digits we'll get the cubic root of the number. The equation is:

$a^{3} = 1000a + 100b + 10c + d$

$a^{3} - 1000a = 100b + 10c + d$

Since 100b + 10c + d is between 0 and 999, that gives us:

$0 < a^{3} - 1000a < 999$

$0 < a(a^{2} - 1000) < 999$

That leads us the following 2 conditions:

$condition 1 : 0 < a(a^{2} - 1000)$

$a(a^{2} - 1000) > 0$

$a^{2} > 1000$

a > 31

$condition 2 : a(a^{2} - 1000) < 999$

$a < 33$

Since 31 < a < 33, therefore a = 32.

$x = 32^{3} = 32768$

$Answer = \boxed {32768}$

Let's consider the reverse problem, find a $y$ such that $y^3$ has the digits of $y$ as its left-most digits, and 3 extra digits. From this we immediately obtain the inequality $y^3 \geq 10^3 y \implies y^2 \geq 10^3 \implies y \geq 32$ .

And equally evident is the inequality $y^3 \leq 10^3(y+1)$ . We can easily solve this in the following way: $y^3 \leq 10^3(y+1) \iff y^3 \leq \sqrt{10^3} \sqrt{10^3} (y+1) \implies y^3 \leq 32 \times 32 (y+1)$ . From this it is clear that if $y > 32$ , the inequality is impossible. Thus we have concluded that $32 \leq y \leq 32$ .

The solution is then $32^3 = 32768$