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 need to analyze the given functions and the conditions provided.

### Step 1: Analyze the function \( f(x) \)
The function is given as:
\[ f(x) = x^3 + ax^2 + bx + c \]
We know that \( f(1) = 0 \). Therefore:
\[ f(1) = 1^3 + a(1^2) + b(1) + c = 1 + a + b + c = 0 \]
This gives us our first equation:
\[ 1 + a + b + c = 0 \quad \text{(Equation 1)} \]

### Step 2: Analyze the function \( g(x) \)
The function is given as:
\[ g(x) = x^3 + bx^2 + cx + a \]
The roots of \( g(x) = 0 \) are the squares of the roots of \( f(x) = 0 \). Let the roots of \( f(x) = 0 \) be \( r_1, r_2, r_3 \). Then the roots of \( g(x) = 0 \) are \( r_1^2, r_2^2, r_3^2 \).

### Step 3: Use Vieta's formulas for \( f(x) \)
From Vieta's formulas, for the cubic polynomial \( f(x) \):
- The sum of the roots \( r_1 + r_2 + r_3 = -a \)
- The sum of the products of the roots taken two at a time \( r_1 r_2 + r_2 r_3 + r_3 r_1 = b \)
- The product of the roots \( r_1 r_2 r_3 = -c \)

### Step 4: Use Vieta's formulas for \( g(x) \)
For the polynomial \( g(x) \):
- The sum of the roots \( r_1^2 + r_2^2 + r_3^2 = -b \)
- The sum of the products of the roots taken two at a time \( r_1^2 r_2^2 + r_2^2 r_3^2 + r_3^2 r_1^2 = c \)
- The product of the roots \( r_1^2 r_2^2 r_3^2 = -a \)

### Step 5: Relate the roots of \( f(x) \) and \( g(x) \)
Using the identity for the sum of squares:
\[ r_1^2 + r_2^2 + r_3^2 = (r_1 + r_2 + r_3)^2 - 2(r_1 r_2 + r_2 r_3 + r_3 r_1) \]
Substituting the values from Vieta's:
\[ r_1^2 + r_2^2 + r_3^2 = (-a)^2 - 2b = a^2 - 2b \]
Setting this equal to the sum of the roots of \( g(x) \):
\[ a^2 - 2b = -b \quad \Rightarrow \quad a^2 - b = 0 \quad \Rightarrow \quad b = a^2 \quad \text{(Equation 2)} \]

### Step 6: Relate the product of the roots
The product of the roots gives:
\[ r_1^2 r_2^2 r_3^2 = (r_1 r_2 r_3)^2 = (-c)^2 = c^2 \]
Setting this equal to the product of the roots of \( g(x) \):
\[ c^2 = -a \quad \text{(Equation 3)} \]

### Step 7: Substitute Equations 2 and 3 into Equation 1
Substituting \( b = a^2 \) into Equation 1:
\[ 1 + a + a^2 + c = 0 \]
Now substituting \( c = -\sqrt{-a} \) (from Equation 3):
\[ 1 + a + a^2 - \sqrt{-a} = 0 \]

### Step 8: Solve for \( a, b, c \)
From Equation 3, we know \( c = -a^2 \). Thus:
\[ 1 + a + a^2 - (-a) = 0 \quad \Rightarrow \quad 1 + 2a + a^2 = 0 \]
This is a quadratic equation:
\[ a^2 + 2a + 1 = 0 \quad \Rightarrow \quad (a + 1)^2 = 0 \quad \Rightarrow \quad a = -1 \]
Substituting \( a = -1 \) back into Equations 2 and 3:
\[ b = (-1)^2 = 1, \quad c = -(-1)^2 = -1 \]

### Step 9: Calculate \( a^{2013} + b^{2013} + c^{2013} \)
Now we have:
\[ a = -1, \quad b = 1, \quad c = -1 \]
Calculating:
\[ a^{2013} + b^{2013} + c^{2013} = (-1)^{2013} + 1^{2013} + (-1)^{2013} = -1 + 1 - 1 = -1 \]

### Final Answer
Thus, the value of \( a^{2013} + b^{2013} + c^{2013} \) is:
\[ \boxed{-1} \]