Let `a ,b ,a n dc` be distinct nonzero real numbers such that `(1-a^3)/a=(1-b^3)/b=(1-c^3)/cdot` The value of `(a^3+b3+c^3)` is _____________.

Explanation

To solve the problem, we need to find the value of \( a^3 + b^3 + c^3 \) given that:

\[ \frac{1 - a^3}{a} = \frac{1 - b^3}{b} = \frac{1 - c^3}{c} = k \]

### Step-by-step Solution:

1. **Set up the equations**:
We start by letting:
\[ \frac{1 - a^3}{a} = k \implies 1 - a^3 = ka \implies a^3 + ka - 1 = 0 \]
Similarly, we can write the equations for \( b \) and \( c \):
\[ b^3 + kb - 1 = 0 \]
\[ c^3 + kc - 1 = 0 \]

2. **Form a polynomial**:
Since \( a, b, c \) are the roots of the polynomial \( x^3 + kx - 1 = 0 \), we can denote this polynomial as:
\[ P(x) = x^3 + kx - 1 \]

3. **Use Vieta's formulas**:
By Vieta's formulas, for a cubic polynomial \( x^3 + px^2 + qx + r = 0 \):
- The sum of the roots \( a + b + c = -p \)
- The sum of the product of the roots taken two at a time \( ab + bc + ca = q \)
- The product of the roots \( abc = -r \)

In our case, since there is no \( x^2 \) term, we have:
\[ a + b + c = 0 \]
\[ ab + ac + bc = k \]
\[ abc = 1 \]

4. **Find \( a^3 + b^3 + c^3 \)**:
We can use the identity:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
Since \( a + b + c = 0 \), this simplifies to:
\[ a^3 + b^3 + c^3 - 3abc = 0 \]
Therefore:
\[ a^3 + b^3 + c^3 = 3abc \]

5. **Substitute the value of \( abc \)**:
We know \( abc = 1 \), so:
\[ a^3 + b^3 + c^3 = 3 \cdot 1 = 3 \]

### Final Answer:
The value of \( a^3 + b^3 + c^3 \) is \( \boxed{3} \).