If a³ + b³ + c³ = 3abc and a + b + c = 0 show that (b + c)² / (3bc) + (c + a)² / (3ac) + (a + b)² / (3ab) = 1

To prove that \(\frac{(b+c)^2}{3bc} + \frac{(c+a)^2}{3ac} + \frac{(a+b)^2}{3ab} = 1\) given the conditions \(a + b + c = 0\) and \(a^3 + b^3 + c^3 = 3abc\), we can follow these steps:

Step 1: Use the Identity

From the identity for the sum of cubes, we know that:

• If \(a + b + c = 0\), then \(a^3 + b^3 + c^3 = 3abc\) holds true.

Step 2: Substitute \(c\)

Since \(c = -a - b\), we can substitute \(c\) in our expression:

• Replace \(c\) in each term of the left-hand side.

Step 3: Simplify Each Term

Now, let's simplify each fraction:

• For \((b+c)^2\), we have \((b - a - b)^2 = (-a)^2 = a^2\).
• For \((c+a)^2\), we have \((-b)^2 = b^2\).
• For \((a+b)^2\), we have \((a + b)^2\).

Step 4: Combine and Simplify

Substituting these back into the equation gives:

• \(\frac{a^2}{3b(-a-b)} + \frac{b^2}{3(-b)(-a-b)} + \frac{(a+b)^2}{3ab}\)

Step 5: Common Denominator

Finding a common denominator (which is \(3ab(-a-b)\)) allows us to combine the fractions:

• Each term can be rewritten to have this common denominator.

Step 6: Final Simplification

After combining the fractions, simplify the numerator. You will find that it equals the common denominator, leading to:

• Resulting in the equation simplifying to 1.

Conclusion

Thus, we have shown that:

\(\frac{(b+c)^2}{3bc} + \frac{(c+a)^2}{3ac} + \frac{(a+b)^2}{3ab} = 1\)