If `a^(1//3)+b^(1//3)+c^(1//3)=0` , then show that `(a+b+c)^(3)=27` abc .

Answer

To solve the problem, we need to show that if \( a^{1/3} + b^{1/3} + c^{1/3} = 0 \), then \( (a + b + c)^3 = 27abc \).

### Step-by-Step Solution:

1. **Let \( x = a^{1/3} \), \( y = b^{1/3} \), and \( z = c^{1/3} \)**:
\[ x + y + z = 0 \]

2. **Using the identity for the sum of cubes**:
The identity states:
\[ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \]
Since \( x + y + z = 0 \), we can simplify this to:
\[ x^3 + y^3 + z^3 = 3xyz \]

3. **Substituting back for \( x, y, z \)**:
We have:
\[ a + b + c = x^3 + y^3 + z^3 \]
Thus,
\[ a + b + c = 3xyz \]
where \( x = a^{1/3}, y = b^{1/3}, z = c^{1/3} \).

4. **Cubing both sides**:
Now, we need to find \( (a + b + c)^3 \):
\[ (a + b + c)^3 = (3xyz)^3 = 27(xyz)^3 \]

5. **Substituting back for \( xyz \)**:
Since \( xyz = a^{1/3} b^{1/3} c^{1/3} = (abc)^{1/3} \), we have:
\[ (xyz)^3 = abc \]
Therefore,
\[ (a + b + c)^3 = 27abc \]

### Conclusion:
Thus, we have shown that if \( a^{1/3} + b^{1/3} + c^{1/3} = 0 \), then \( (a + b + c)^3 = 27abc \).