AMC 10 · 2025 · #19
Grade 11 algebraPick an answer.
The roots a, b, c are never found one by one, and they don't need to be. Any expression symmetric in the roots can be rewritten using the three elementary symmetric quantities e1 = a+b+c, e2 = ab+bc+ca, and e3 = abc, and Vieta's formulas read those three numbers straight off the polynomial's coefficients. So the plan is to name those quantities, use the structure of S to fold it down until only e1, e2, e3 remain, then substitute their known values. The lucky break that makes the folding short is e1 = 0, which collapses each grouped binomial in one stroke.
Expand the factored cubic
Since a, b, c are the roots, the cubic equals (x-a)(x-b)(x-c); expanding gives x³ - (a+b+c)x² + (ab+bc+ca)x - abc.
A polynomial with known roots is just the product of its linear factors, so multiplying them out shows exactly how the roots build the coefficients.
9.A-APR.A.1Introduce A VariableRead off Vieta's relations
Matching coefficients with x³ + 0x² + kx + 1 gives a+b+c = 0, ab+bc+ca = k, abc = -1 — the only root facts we need.
Two ways of writing the same polynomial must agree coefficient by coefficient, which hands you the root sums for free.
Two ways of writing the same polynomial must agree coefficient by coefficient, which hands over the root sums.
▸ Why?
When two expressions are equal for every input, each matching coefficient must be equal too.
▸ Why?
Those coefficients record exactly the symmetric combinations of the roots and nothing else.
Group by the shared square
Pair the six terms by their shared square, so S = a²b²(a+b) + b²c²(b+c) + c²a²(c+a).
Pulling out the common square turns a shapeless six-term sum into three neat pieces tied to the root sums.
9.A-SSE.A.2Organize Information In More WaysUse a+b+c = 0
Since a+b+c = 0, each binomial is minus the missing root (a+b = -c, etc.), so S = -(a²b²c + ab²c² + a²bc²).
When the three roots add to zero, any two of them add up to minus the third, which erases the binomials in one move.
9.A-SSE.A.2Organize Information In More WaysFactor out abc
Each term hides one a, one b, one c, so pulling out abc leaves S = -abc(ab + bc + ca) — both known Vieta quantities.
Every term hides one full copy of the product abc, so factoring it out reveals the familiar symmetric pieces underneath.
9.A-SSE.A.2Organize Information In More WaysSubstitute the known values
Plug in abc = -1 and ab+bc+ca = k: S = -(-1)(k) = k — choice (E).
Once the expression is written with only e2 and e3, substituting their coefficient values finishes it instantly.
11.A-APR.C.4Introduce A VariableA messy symmetric sum in the roots is never as scary as it looks: rewrite it using a+b+c, ab+bc+ca, and abc, which you read straight off the coefficients, and here it melts down to just k.
- Expand the factored cubic
- Read off Vieta's relations
- Group by the shared square
- Use a+b+c = 0
- Factor out abc
- Substitute the known values