In short
Whenever you see "find a^3 + b^3 + c^3 given a + b + c = 0", do not expand. Reach for the identity
The right-hand side has (a + b + c) as a factor. With a + b + c = 0, the entire RHS vanishes, so
That is the whole trick. One line. No cubing, no expansion, no algebra grind.
You are sitting in a JEE Advanced paper. The problem says: "If a + b + c = 0 and abc = 7, find a^3 + b^3 + c^3." The slow student starts trying to solve for a, b, c individually. They will not finish in time. The fast student writes "3 \times 7 = 21" and moves on.
The difference is one identity, used as a recognition rule rather than a computation.
The recognition rule
The signal you are looking for is the phrase "a + b + c = 0" (or any equivalent — three roots of a depressed cubic, three numbers whose sum vanishes, three vectors whose components sum to zero, etc.) sitting next to a request for a^3 + b^3 + c^3 or a^3 + b^3 + c^3 - 3abc or anything in that family.
The moment you see that signal, do not think. Apply the identity:
Plug in a + b + c = 0:
Why a + b + c = 0 makes the second factor irrelevant: the second factor (a^2 + b^2 + c^2 - ab - bc - ca) could be any number — positive, negative, complicated, full of cross terms. It does not matter. It is being multiplied by zero. Whatever it is, the product is zero. You never need to compute the second factor when the first one vanishes.
So a^3 + b^3 + c^3 = 3abc. That is the punchline.
Why this pattern is so common in JEE and Olympiad problems: the condition a + b + c = 0 is a clean algebraic constraint — it ties three unknowns together with a single linear equation, leaving two degrees of freedom. The cube-sum a^3 + b^3 + c^3 is a symmetric expression in those three unknowns. Symmetry plus the constraint forces the answer to depend on a single invariant (abc in this case). Problem-setters love this because it tests whether you spot the structure rather than grinding through the algebra. One identity, one parameter to compute.
Three worked examples
Example 1: $a + b + c = 0$ and $abc = 5$, find $a^3 + b^3 + c^3$
You do not need to know a, b, or c individually. You do not need to expand anything. Apply the collapsed form directly:
Why this works without knowing a, b, c: the identity a^3 + b^3 + c^3 - 3abc = (a+b+c)(\ldots) holds for every triple (a, b, c). When (a + b + c) = 0, the right-hand side is forced to zero regardless of what the individual values are. So a^3 + b^3 + c^3 is locked to 3abc — and you only need the product, not the individual numbers.
Result. a^3 + b^3 + c^3 = 15. One multiplication, no expansion.
Example 2: $x = 4$, $y = -3$, $z = -1$ — verify the collapse directly
First check the trigger: x + y + z = 4 + (-3) + (-1) = 0. The condition holds.
Compute the cube-sum directly:
Compute 3xyz:
Both equal 36. The collapse is verified — x^3 + y^3 + z^3 = 3xyz exactly when x + y + z = 0.
Why the signs work out: 4^3 = 64 is positive, (-3)^3 = -27 and (-1)^3 = -1 are negative. On the other side, 4 \cdot (-3) \cdot (-1) = 12 — two negatives multiplied together flip back to positive. The arithmetic looks like a coincidence but is forced by the identity.
Result. Both sides equal 36. The identity holds.
Example 3: factor $a^3 + b^3 + c^3 - 3abc$ when $a + b + c = 0$
Apply the identity:
With a + b + c = 0:
So the whole expression a^3 + b^3 + c^3 - 3abc is identically zero whenever the sum of the variables is zero. This means a^3 + b^3 + c^3 and 3abc are not just equal in value — they are the same expression once the constraint is imposed.
Why this is useful for olympiad problems: it lets you replace any occurrence of a^3 + b^3 + c^3 in a complicated expression with 3abc, or vice versa, as long as the sum-zero condition holds. That swap often turns a degree-3 symmetric mess into a simple product, which is much easier to bound or estimate.
Result. Under a + b + c = 0, the expression a^3 + b^3 + c^3 - 3abc is identically zero — so the two cube-quantities are interchangeable.
Why this single shortcut saves 2–3 minutes per problem
A JEE Advanced paper gives you about three minutes per question. Without the identity, a problem like Example 1 looks like a system of equations: you would try to find a, b, c individually from the given constraints, cube them, sum them. That path is hopeless — the constraints do not pin down the individual values, only the symmetric combinations.
With the identity, the problem collapses to a single multiplication. You write the answer in five seconds and move on.
This is not just for cube-sum problems. Recognition of the pattern lets you tackle:
- "Roots of x^3 + px + q = 0 — find \alpha^3 + \beta^3 + \gamma^3" (the depressed cubic has \alpha + \beta + \gamma = 0 from Vieta's, so the answer is 3\alpha\beta\gamma = -3q).
- "Three vectors \vec a, \vec b, \vec c with \vec a + \vec b + \vec c = \vec 0" (componentwise, the cube-sum collapse applies to each coordinate).
- "Prove x^3 + y^3 + z^3 = 3xyz given x + y + z = 0" (one line, just cite the identity).
The identity is famous. It appears in Hall and Knight, in Higher Algebra, in every JEE preparation book, and on the back cover of every Indian school's algebra cheat-sheet. Knowing it is not optional.
When the sum is not zero
If a + b + c \neq 0, the second factor stops being irrelevant. You then need the full identity:
Often the second factor is computed indirectly, using
So if a problem gives you a + b + c = s and ab + bc + ca = p and abc = q, you can write
That is the "full" cube-sum formula in terms of elementary symmetric polynomials. When s = 0, the second term vanishes and you are back to 3q — the collapse case.
The mental shortcut is: see "a + b + c = 0" → write 3abc. See "a + b + c \neq 0" → use the full formula. Two patterns, two reflexes.
Connection to the bigger identity
This recognition rule is the most-used corollary of the beautiful identity, which gives the full factorisation of x^3 + y^3 + z^3 - 3xyz. The parent identity is the structural truth; this satellite is the one tactic you actually deploy in exams.
If you want the geometric and algebraic intuition for why (x + y + z) is forced to be a factor in the first place, see the parent. If you just want the shortcut for the next JEE problem, the rule above is enough — see "a + b + c = 0" and write 3abc.
References
- Factor Theorem (Wikipedia) — explains why (a+b+c) divides a^3+b^3+c^3-3abc.
- Vieta's Formulas (Wikipedia) — the source of "sum of roots = 0" in depressed cubics.
- Symmetric Polynomial (Wikipedia) — power sums like a^3+b^3+c^3 in terms of elementary symmetric polynomials.
- Newton's Identities (Wikipedia) — generalises this collapse to higher power sums.