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The problem states that real numbers \(a\), \(b\), and \(c\) satisfy the conditions:

\[
abc = -1, \quad a + b + c = 4
\]

It asks to find \(a^2 + b^2 + c^2\) based on the following equation:

\[
\frac{a}{a^2 - 3a - 1} + \frac{b}{b^2 - 3b - 1} + \frac{c}{c^2 - 3c - 1} = \frac{4}{9}
\]

Let's begin by analyzing the given information and solving for \(a^2 + b^2 + c^2\). I'll work through the details of simplifying the equation and incorporating the conditions provided.We have the expanded form of \((a + b + c)^2\):

\[
16 = a^2 + b^2 + c^2 + 2(ab + bc + ca)
\]

Next, we need to express \(ab + bc + ca\). By using the symmetric identities derived from the sum and product of roots in cubic equations, we'll substitute these into the equation to isolate \(a^2 + b^2 + c^2\). Let me continue from here.The expression for \( ab + bc + ca \) is:

\[
ab + bc + ca = -\frac{1}{2}(a^2 + b^2 + c^2) + 8
\]

Now, we can substitute this result back into the equation \( 16 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \) and solve for \( a^2 + b^2 + c^2 \). Let me do that next.

Given that real numbers a, b, c satisfy abc = -1 and a + b + c = 4, find the value of a^2 + b^2 + c^2 using the equation: a / (a^2 - 3a - 1) + b / (b^2 - 3b - 1) + c / (c^2 - 3c - 1) = 4/9.

This problem involves solving for a^2 + b^2 + c^2 where a, b, and c satisfy abc = -1 and a + b + c = 4. The equation involves symmetric identities and requires algebraic manipulation. Suitable for advanced high school students.

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