`2sqrt(2)a^3+3sqrt(3)b^3+c^3-3sqrt(6)abc`

Answer

To factor the polynomial \(2\sqrt{2}a^3 + 3\sqrt{3}b^3 + c^3 - 3\sqrt{6}abc\), we can use the identity for the sum of cubes. The identity states:

\[ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \]

### Step-by-Step Solution:

1. **Identify the terms**: We have the expression \(2\sqrt{2}a^3 + 3\sqrt{3}b^3 + c^3 - 3\sqrt{6}abc\).

2. **Rewrite the terms**: We can express \(2\sqrt{2}a^3\) as \((\sqrt[3]{2\sqrt{2}}a)^3\), \(3\sqrt{3}b^3\) as \((\sqrt[3]{3\sqrt{3}}b)^3\), and \(c^3\) remains as is. We need to express \(-3\sqrt{6}abc\) in a similar form.

3. **Express \(-3\sqrt{6}abc\)**: We can rewrite \(-3\sqrt{6}abc\) as \(-3(\sqrt{2}a)(\sqrt{3}b)(c)\).

4. **Assign variables**: Let:
- \(x = \sqrt[3]{2\sqrt{2}}a\)
- \(y = \sqrt[3]{3\sqrt{3}}b\)
- \(z = c\)

5. **Apply the identity**: Now we can apply the identity:
\[ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \]

6. **Calculate \(x + y + z\)**:
\[ x + y + z = \sqrt[3]{2\sqrt{2}}a + \sqrt[3]{3\sqrt{3}}b + c \]

7. **Calculate \(x^2 + y^2 + z^2\)**:
\[ x^2 = (\sqrt[3]{2\sqrt{2}}a)^2 = \frac{2\sqrt{2}}{3}a^2 \]
\[ y^2 = (\sqrt[3]{3\sqrt{3}}b)^2 = \frac{3\sqrt{3}}{3}b^2 \]
\[ z^2 = c^2 \]

8. **Calculate \(xy, yz, zx\)**:
\[ xy = \sqrt[3]{2\sqrt{2}}a \cdot \sqrt[3]{3\sqrt{3}}b = \sqrt[3]{6\sqrt{6}}ab \]
\[ yz = \sqrt[3]{3\sqrt{3}}b \cdot c \]
\[ zx = \sqrt[3]{2\sqrt{2}}a \cdot c \]

9. **Combine the results**: Substitute these values into the identity to get the factorization:
\[ (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \]

10. **Final expression**: The final factorized form will be:
\[ \left(\sqrt[3]{2\sqrt{2}}a + \sqrt[3]{3\sqrt{3}}b + c\right) \left(\text{remaining terms}\right) \]