Let `f(x) = x^3 + ax^2 + bx + c` and `g(x) = x^3 + bx^2 + cx + a`, where a, b, c are integers. Suppose that the following conditions hold-
(a) `f(1) = 0`,
(b) the roots of g(x) = 0 are the squares of the roots of f(x) = 0. Find the value of: `a^(2013) + b^(2013) + c^(2013)`

Answer

To solve the problem, we will follow these steps:

1. **Understanding the Functions**:
We have two cubic functions:
\[ f(x) = x^3 + ax^2 + bx + c \]
\[ g(x) = x^3 + bx^2 + cx + a \]
where \( a, b, c \) are integers.

2. **Using the Given Conditions**:
- From the condition \( f(1) = 0 \):
\[ f(1) = 1 + a + b + c = 0 \implies a + b + c = -1 \quad \text{(1)} \]
- The roots of \( g(x) = 0 \) are the squares of the roots of \( f(x) = 0 \). Let the roots of \( f(x) = 0 \) be \( 1, p, q \). Then the roots of \( g(x) = 0 \) are \( 1^2, p^2, q^2 \).

3. **Finding the Sum and Product of Roots**:
- For \( f(x) \):
\[ 1 + p + q = -a \quad \text{(sum of roots)} \]
\[ 1pq + p + q = b \quad \text{(sum of products of roots taken two at a time)} \]
\[ 1pq = -c \quad \text{(product of roots)} \]
- For \( g(x) \):
\[ 1 + p^2 + q^2 = -b \quad \text{(sum of roots)} \]
\[ 1p^2 + 1q^2 + p^2q^2 = c \quad \text{(sum of products of roots taken two at a time)} \]
\[ 1p^2q^2 = -a \quad \text{(product of roots)} \]

4. **Relating the Roots**:
- From the sum of roots of \( f(x) \):
\[ p + q = -a - 1 \quad \text{(2)} \]
- From the product of roots of \( f(x) \):
\[ pq = -\frac{c}{1} = -c \quad \text{(3)} \]

5. **Expressing \( p^2 + q^2 \)**:
Using the identity \( p^2 + q^2 = (p + q)^2 - 2pq \):
\[ p^2 + q^2 = (-a - 1)^2 - 2(-c) = (a + 1)^2 + 2c \quad \text{(4)} \]

6. **Substituting into \( g(x) \)**:
- From \( g(x) \):
\[ 1 + p^2 + q^2 = -b \implies 1 + (a + 1)^2 + 2c = -b \quad \text{(5)} \]
- From the product of roots of \( g(x) \):
\[ 1p^2q^2 = -a \implies pq^2 = -a \quad \text{(6)} - Substitute \( pq = -c \) into \( (pq)^2 \): \[ (-c)^2 = c^2 \]

7. **Solving the System**:
- From (1), (5), and (6), we can express \( a, b, c \) in terms of each other and solve for integer solutions.
- After substituting and simplifying, we find:
\[ c = -1, a = -1, b = 1 \]

8. **Final Calculation**:
We need to find \( a^{2013} + b^{2013} + c^{2013} \):
\[ a^{2013} + b^{2013} + c^{2013} = (-1)^{2013} + 1^{2013} + (-1)^{2013} = -1 + 1 - 1 = -1 \]

Thus, the final answer is:
\[ \boxed{-1} \]