Solving the Algebra Problem: (a + b + c = 0)

We are given a condition in this algebra problem: \(a + b + c = 0\). We need to find the value of the expression \((a^3 + b^3 + c^3)^2\).

Understanding the Key Algebra Identity

There is a very important algebraic identity that relates the sum of cubes to the sum of the variables. The general identity is:

\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \]

This identity is crucial for solving problems like this.

Applying the Condition a + b + c = 0

Now, let's use the given condition that \(a + b + c = 0\). We substitute this into the identity:

\[ a^3 + b^3 + c^3 - 3abc = (\mathbf{0})(a^2 + b^2 + c^2 - ab - bc - ca) \]

Since anything multiplied by zero is zero, the right side of the equation becomes zero:

\[ a^3 + b^3 + c^3 - 3abc = 0 \]

Rearranging this equation, we get the specific identity that applies when the sum of the variables is zero:

\[ a^3 + b^3 + c^3 = 3abc \]

This is the key result we need when \(a + b + c = 0\).

Evaluating the Expression (a³ + b³ + c³)²

We are asked to find the value of \((a^3 + b^3 + c^3)^2\). Using the identity we just derived (\(a^3 + b^3 + c^3 = 3abc\) when \(a + b + c = 0\)), we can substitute \(3abc\) for \((a^3 + b^3 + c^3)\):

\[ (a^3 + b^3 + c^3)^2 = (3abc)^2 \]

Now, we need to square the term \(3abc\). Squaring a product means squaring each factor in the product:

\[ (3abc)^2 = 3^2 \cdot a^2 \cdot b^2 \cdot c^2 \]

Calculating \(3^2\):

\[ 3^2 = 9 \]

So, the expression simplifies to:

\[ (3abc)^2 = 9a^2b^2c^2 \]

Final Result

Therefore, if \(a + b + c = 0\), the value of \((a^3 + b^3 + c^3)^2\) is \(9a^2b^2c^2\).

Step-by-Step Summary

1. Start with the given condition: \(a + b + c = 0\).
2. Recall or derive the identity: If \(a + b + c = 0\), then \(a^3 + b^3 + c^3 = 3abc\).
3. Substitute this result into the required expression: \((a^3 + b^3 + c^3)^2\).
4. The expression becomes \((3abc)^2\).
5. Square the term: \((3abc)^2 = 3^2 \cdot a^2 \cdot b^2 \cdot c^2 = 9a^2b^2c^2\).

Revision Table: Key Identities

Additional Information: Why is a³ + b³ + c³ = 3abc when a + b + c = 0?

Let's briefly see why the identity \(a^3 + b^3 + c^3 = 3abc\) holds true when \(a+b+c=0\).

Given \(a+b+c=0\), we can write \(c = -(a+b)\). Substitute this into \(c^3\):

\[ c^3 = (-(a+b))^3 = -(a+b)^3 \]

Recall the binomial expansion: \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\). So,

\[ c^3 = -(a^3 + 3a^2b + 3ab^2 + b^3) = -a^3 - 3a^2b - 3ab^2 - b^3 \]

Now consider \(a^3 + b^3 + c^3\). Substitute the expression for \(c^3\):

\[ a^3 + b^3 + c^3 = a^3 + b^3 + (-a^3 - 3a^2b - 3ab^2 - b^3) \]

\[ a^3 + b^3 + c^3 = a^3 + b^3 - a^3 - 3a^2b - 3ab^2 - b^3 \]

The \(a^3\) terms cancel, and the \(b^3\) terms cancel:

\[ a^3 + b^3 + c^3 = -3a^2b - 3ab^2 \]

Factor out \(-3ab\) from the right side:

\[ a^3 + b^3 + c^3 = -3ab(a + b) \]

From the condition \(a + b + c = 0\), we know that \(a + b = -c\). Substitute this back into the equation:

\[ a^3 + b^3 + c^3 = -3ab(\mathbf{-c}) \]

\[ a^3 + b^3 + c^3 = 3abc \]

This derivation confirms the identity used to solve the problem. Understanding this derivation helps solidify why the identity holds under the given condition \(a + b + c = 0\).