Question: Let $ a, b, c $ be real numbers such that $ a + b + c = 0 $. Prove that $ a^3 + b^3 + c^3 = 3abc $. - ToelettAPP
Proving $ a^3 + b^3 + c^3 = 3abc $ Given $ a + b + c = 0 $
Proving $ a^3 + b^3 + c^3 = 3abc $ Given $ a + b + c = 0 $
Introduction
In algebra, one of the most elegant identities involving symmetric sums of three variables $ a, b, c $ is the identity:
$$
a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)
$$
This identity holds for all real numbers $ a, b, c $. However, a particularly useful special case arises when $ a + b + c = 0 $. In this article, we will prove that if $ a + b + c = 0 $, then
$$
a^3 + b^3 + c^3 = 3abc
$$
through direct algebraic manipulation, using the given condition to simplify the expression.
Step-by-Step Proof
Understanding the Context
Start with the well-known algebraic identity valid for any real numbers $ a, b, c $:
$$
a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)
$$
This identity can be verified by expanding the right-hand side, but for now, we accept it as a standard result.
Given in the problem is the condition:
$$
a + b + c = 0
$$
Substitute this into the factor on the right-hand side:
$$
a^3 + b^3 + c^3 - 3abc = (0)(a^2 + b^2 + c^2 - ab - bc - ca) = 0
$$
Therefore,
$$
a^3 + b^3 + c^3 - 3abc = 0
$$
Adding $ 3abc $ to both sides yields:
$$
a^3 + b^3 + c^3 = 3abc
$$
Alternative Derivation for Deeper Understanding
To strengthen conceptual understanding, consider expressing one variable in terms of the others using $ c = -a - b $. Substitute into $ a^3 + b^3 + c^3 $:
$$
c = -a - b
$$
$$
c^3 = (-a - b)^3 = - (a + b)^3 = - (a^3 + 3a^2b + 3ab^2 + b^3)
$$
Now compute $ a^3 + b^3 + c^3 $:
$$
a^3 + b^3 + c^3 = a^3 + b^3 - (a^3 + 3a^2b + 3ab^2 + b^3) = -3a^2b - 3ab^2
$$
Factor out $ -3ab $:
$$
a^3 + b^3 + c^3 = -3ab(a + b)
$$
But since $ c = -a - b $, then $ a + b = -c $, so:
$$
a^3 + b^3 + c^3 = -3ab(-c) = 3abc
$$
Key Insights
This confirms the identity directly using substitution, aligning with the earlier factorization.
Conclusion
We have shown rigorously that when $ a + b + c = 0 $, the identity
$$
a^3 + b^3 + c^3 = 3abc
$$
holds true. This result is particularly useful in simplifying symmetric expressions in algebraic problems involving dependent variables. It exemplifies how a single condition can dramatically simplify expressions, and it is a cornerstone in symmetric polynomial theory.
Whether used in competitive mathematics, calculus, or linear algebra, mastering this identity enhances problem-solving agility—especially in Olympiad-level problems involving sum-zero constraints.
Keywords:
$ a + b + c = 0 $, $ a^3 + b^3 + c^3 $, $ 3abc $, algebraic identity, symmetry in polynomials, real numbers, mathematical proof, Olympiad algebra, identity derivation, substitution method, Olympiad mathematics.
Meta Description:
Prove that $ a^3 + b^3 + c^3 = 3abc $ when $ a + b + c = 0 $ using the identity $ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) $.
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Topics: Algebra, Identities, Olympiad Mathematics, Symmetric Sums, Real Numbers
