Aida made three cubes with positive integer side lengths $a,b,c$. They were too small for her, so she divided them into unit cubes and attempted to construct a cube of side $a+b+c$. Unfortunately, she was $648$ blocks off. How many possibilities of the ordered triple $\left(a,b,c\right)$ are there?

[i]2017 CCA Math Bonanza Team Round #9[/i]
We start with the given equation:
\[ a^3 + b^3 + c^3 + 648 = (a + b + c)^3 \]

First, we expand the right-hand side:
\[ (a + b + c)^3 = a^3 + b^3 + c^3 + 3a^2b + 3a^2c + 3b^2a + 3b^2c + 3c^2a + 3c^2b + 6abc \]

Subtracting \(a^3 + b^3 + c^3\) from both sides, we get:
\[ 648 = 3a^2b + 3a^2c + 3b^2a + 3b^2c + 3c^2a + 3c^2b + 6abc \]

Dividing both sides by 3:
\[ 216 = a^2b + a^2c + b^2a + b^2c + c^2a + c^2b + 2abc \]

We will now consider different values for \(a\), \(b\), and \(c\) under the assumption \(a \leq b \leq c\).

### Case 1: \(a = b = c\)
If \(a = b = c\), then:
\[ 3a^3 + 648 = (3a)^3 \]
\[ 3a^3 + 648 = 27a^3 \]
\[ 648 = 24a^3 \]
\[ a^3 = 27 \]
\[ a = 3 \]

Thus, \((a, b, c) = (3, 3, 3)\) is a solution.

### Case 2: \(a = 2\)
Substituting \(a = 2\) into the equation:
\[ 216 = 4b + 4c + 2b^2 + 2c^2 + b^2c + c^2b + 4bc \]

This equation is quite complex, so we will try different values for \(b\) and solve for \(c\).

#### Subcase: \(b = 2\)
\[ 216 = 4(2) + 4c + 2(2^2) + 2c^2 + 2^2c + c^2(2) + 4(2)c \]
\[ 216 = 8 + 4c + 8 + 2c^2 + 4c + 2c^2 + 8c \]
\[ 216 = 16 + 4c + 2c^2 + 4c + 2c^2 + 8c \]
\[ 216 = 16 + 14c + 4c^2 \]
\[ 200 = 14c + 4c^2 \]
\[ 4c^2 + 14c - 200 = 0 \]

Solving this quadratic equation:
\[ c = \frac{-14 \pm \sqrt{14^2 - 4 \cdot 4 \cdot (-200)}}{2 \cdot 4} \]
\[ c = \frac{-14 \pm \sqrt{196 + 3200}}{8} \]
\[ c = \frac{-14 \pm \sqrt{3396}}{8} \]

Since \(\sqrt{3396}\) is not an integer, there are no integer solutions for \(c\) in this subcase.

### Case 3: \(a = 1\)
Substituting \(a = 1\) into the equation:
\[ 216 = b + c + b^2 + c^2 + b^2c + c^2b + 2bc \]

#### Subcase: \(b = 1\)
\[ 216 = 1 + c + 1 + c^2 + c + c^2 + 2c \]
\[ 216 = 2 + 4c + 2c^2 \]
\[ 214 = 4c + 2c^2 \]
\[ 2c^2 + 4c - 214 = 0 \]

Solving this quadratic equation:
\[ c = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 2 \cdot (-214)}}{2 \cdot 2} \]
\[ c = \frac{-4 \pm \sqrt{16 + 1712}}{4} \]
\[ c = \frac{-4 \pm \sqrt{1728}}{4} \]

Since \(\sqrt{1728}\) is not an integer, there are no integer solutions for \(c\) in this subcase.

#### Subcase: \(b = 2\)
\[ 216 = 2 + c + 4 + c^2 + 4c + c^2 + 2c \]
\[ 216 = 6 + 7c + 2c^2 \]
\[ 210 = 7c + 2c^2 \]
\[ 2c^2 + 7c - 210 = 0 \]

Solving this quadratic equation:
\[ c = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 2 \cdot (-210)}}{2 \cdot 2} \]
\[ c = \frac{-7 \pm \sqrt{49 + 1680}}{4} \]
\[ c = \frac{-7 \pm \sqrt{1729}}{4} \]

Since \(\sqrt{1729}\) is not an integer, there are no integer solutions for \(c\) in this subcase.

#### Subcase: \(b = 3\)
\[ 216 = 3 + c + 9 + c^2 + 9c + c^2 + 2c \]
\[ 216 = 12 + 12c + 2c^2 \]
\[ 204 = 12c + 2c^2 \]
\[ 2c^2 + 12c - 204 = 0 \]

Solving this quadratic equation:
\[ c = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 2 \cdot (-204)}}{2 \cdot 2} \]
\[ c = \frac{-12 \pm \sqrt{144 + 1632}}{4} \]
\[ c = \frac{-12 \pm \sqrt{1776}}{4} \]

Since \(\sqrt{1776}\) is not an integer, there are no integer solutions for \(c\) in this subcase.

#### Subcase: \(b = 4\)
\[ 216 = 4 + c + 16 + c^2 + 16c + c^2 + 2c \]
\[ 216 = 20 + 18c + 2c^2 \]
\[ 196 = 18c + 2c^2 \]
\[ 2c^2 + 18c - 196 = 0 \]

Solving this quadratic equation:
\[ c = \frac{-18 \pm \sqrt{18^2 - 4 \cdot 2 \cdot (-196)}}{2 \cdot 2} \]
\[ c = \frac{-18 \pm \sqrt{324 + 1568}}{4} \]
\[ c = \frac{-18 \pm \sqrt{1892}}{4} \]

Since \(\sqrt{1892}\) is not an integer, there are no integer solutions for \(c\) in this subcase.

#### Subcase: \(b = 5\)
\[ 216 = 5 + c + 25 + c^2 + 25c + c^2 + 2c \]
\[ 216 = 30 + 27c + 2c^2 \]
\[ 186 = 27c + 2c^2 \]
\[ 2c^2 + 27c - 186 = 0 \]

Solving this quadratic equation:
\[ c = \frac{-27 \pm \sqrt{27^2 - 4 \cdot 2 \cdot (-186)}}{2 \cdot 2} \]
\[ c = \frac{-27 \pm \sqrt{729 + 1488}}{4} \]
\[ c = \frac{-27 \pm \sqrt{2217}}{4} \]

Since \(\sqrt{2217}\) is not an integer, there are no integer solutions for \(c\) in this subcase.

### Conclusion
The only valid solution is \((a, b, c) = (3, 3, 3)\). Since this is the only solution, there is only one ordered triple.

The final answer is \(\boxed{1}\)