AlgebraDifficulty 40-85511 tagged problems

Symmetric Polynomials

Symmetric Polynomials is the habit of using expressions unchanged by variable swaps to reduce variables and exploit root relationships. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.

Practice this technique Match my level

The Key Result

$$p + q = e_1,\quad pq = e_2 \;\implies\; p^2 + q^2 = e_1^2 - 2e_2,\quad p^3 + q^3 = e_1^3 - 3e_1 e_2.$$

Any expression that is unchanged when you swap the variables can be rewritten using only their elementary symmetric sums (the sum and the product), collapsing many variables into a few.

Why It Works

1
Let $e_1 = p + q$ and $e_2 = pq$ be the elementary symmetric polynomials of $p, q$.
2
Square the sum: $(p + q)^2 = p^2 + 2pq + q^2$, so $p^2 + q^2 = e_1^2 - 2e_2$.
3
Cube the sum: $(p + q)^3 = p^3 + 3p^2 q + 3pq^2 + q^3 = p^3 + q^3 + 3pq(p + q)$.
4
Rearrange: $p^3 + q^3 = (p+q)^3 - 3pq(p+q) = e_1^3 - 3e_1 e_2$. Every symmetric polynomial reduces to $e_1, e_2$ this way (Newton's identities generalize it).

Worked Examples

Example 1

If $p + q = 4$ and $pq = 3$, find $p^3 + q^3$.

1
Identify the symmetric sums: $e_1 = p + q = 4$ and $e_2 = pq = 3$.
2
Apply the identity $p^3 + q^3 = e_1^3 - 3e_1 e_2$.
3
Compute: $4^3 - 3(4)(3) = 64 - 36 = 28$.

Answer:

$p^3 + q^3 = 28$.

Warm-up

If $p + q = 6$ and $pq = 8$, find $p^2 + q^2$.

1
Use $p^2 + q^2 = (p + q)^2 - 2pq$.
2
Substitute the symmetric sums: $6^2 - 2(8)$.
3
Compute: $36 - 16 = 20$.

Answer:

$p^2 + q^2 = 20$.

Contest level

Given $x + y = 5$ and $x^2 + y^2 = 13$, find $x^3 + y^3$.

1
First recover the product: $xy = \dfrac{(x + y)^2 - (x^2 + y^2)}{2} = \dfrac{25 - 13}{2} = 6$.
2
Use $x^3 + y^3 = (x + y)^3 - 3xy(x + y)$.
3
Substitute: $5^3 - 3(6)(5) = 125 - 90 = 35$.

Answer:

$x^3 + y^3 = 35$.

Olympiad / Challenge

Real numbers satisfy $x + y + z = 1$, $x^2 + y^2 + z^2 = 2$, and $x^3 + y^3 + z^3 = 3$. Find $x^4 + y^4 + z^4$.

1
Let $e_1, e_2, e_3$ be the elementary symmetric sums and $p_k$ the power sums. Then $e_1 = p_1 = 1$. From $p_2 = e_1 p_1 - 2e_2$: $2 = 1 - 2e_2$, so $e_2 = -\tfrac{1}{2}$.
2
From $p_3 = e_1 p_2 - e_2 p_1 + 3e_3$: $3 = (1)(2) - (-\tfrac12)(1) + 3e_3 = 2 + \tfrac12 + 3e_3$, so $e_3 = \tfrac{1}{6}$.
3
For three variables, $p_4 = e_1 p_3 - e_2 p_2 + e_3 p_1$ (no $e_4$ term): $p_4 = (1)(3) - (-\tfrac12)(2) + (\tfrac16)(1) = 3 + 1 + \tfrac16 = \tfrac{25}{6}$.

Answer:

$x^4 + y^4 + z^4 = \dfrac{25}{6}$.

Going Deeper

Generalization: the Fundamental Theorem of Symmetric Polynomials says EVERY symmetric polynomial is a unique polynomial in the elementary symmetric sums $e_1, e_2, \dots, e_n$; Newton's identities give the explicit recurrence converting between the $e_k$ and the power sums $p_k$.
Where it appears: AIME problems that give a few symmetric conditions and ask for another symmetric quantity; combined with Vieta it evaluates $\sum r_i^k$ for a polynomial's roots without ever finding them.
Pitfall: the conversion identities are degree-specific — $p^3 + q^3 = e_1^3 - 3e_1 e_2$ is for TWO variables; the three-variable analogue $\sum x^3 = e_1^3 - 3e_1 e_2 + 3e_3$ has an extra $+3e_3$. Using the two-variable formula on three variables silently drops that term.

Spot the Signal

Look for problems where the key step is using expressions unchanged by variable swaps to reduce variables and exploit root relationships.
You can describe the hard part as using expressions unchanged by variable swaps to reduce variables and exploit root relationships, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the expression or equation that calls for symmetric polynomials, then rewrite the givens around it.
Name the relation that makes Symmetric Polynomials legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Symmetric Polynomials just because the surface looks familiar; verify the required condition first.
Applying Symmetric Polynomials because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let symmetric polynomials reveal the algebraic structure; then compute only what remains.

Try it on:

Practice a contest problem where the key step is using expressions unchanged by variable swaps to reduce variables and exploit root relationships.

Back to techniques

AlgebraDifficulty 40-85511 tagged problems

Symmetric Polynomials

Practice this technique Match my level

The Key Result

$$p + q = e_1,\quad pq = e_2 \;\implies\; p^2 + q^2 = e_1^2 - 2e_2,\quad p^3 + q^3 = e_1^3 - 3e_1 e_2.$$

Any expression that is unchanged when you swap the variables can be rewritten using only their elementary symmetric sums (the sum and the product), collapsing many variables into a few.

Why It Works

1
Let $e_1 = p + q$ and $e_2 = pq$ be the elementary symmetric polynomials of $p, q$.
2
Square the sum: $(p + q)^2 = p^2 + 2pq + q^2$, so $p^2 + q^2 = e_1^2 - 2e_2$.
3
Cube the sum: $(p + q)^3 = p^3 + 3p^2 q + 3pq^2 + q^3 = p^3 + q^3 + 3pq(p + q)$.
4
Rearrange: $p^3 + q^3 = (p+q)^3 - 3pq(p+q) = e_1^3 - 3e_1 e_2$. Every symmetric polynomial reduces to $e_1, e_2$ this way (Newton's identities generalize it).

Worked Examples

Example 1

If $p + q = 4$ and $pq = 3$, find $p^3 + q^3$.

1
Identify the symmetric sums: $e_1 = p + q = 4$ and $e_2 = pq = 3$.
2
Apply the identity $p^3 + q^3 = e_1^3 - 3e_1 e_2$.
3
Compute: $4^3 - 3(4)(3) = 64 - 36 = 28$.

Answer:

$p^3 + q^3 = 28$.

Warm-up

If $p + q = 6$ and $pq = 8$, find $p^2 + q^2$.

1
Use $p^2 + q^2 = (p + q)^2 - 2pq$.
2
Substitute the symmetric sums: $6^2 - 2(8)$.
3
Compute: $36 - 16 = 20$.

Answer:

$p^2 + q^2 = 20$.

Contest level

Given $x + y = 5$ and $x^2 + y^2 = 13$, find $x^3 + y^3$.

1
First recover the product: $xy = \dfrac{(x + y)^2 - (x^2 + y^2)}{2} = \dfrac{25 - 13}{2} = 6$.
2
Use $x^3 + y^3 = (x + y)^3 - 3xy(x + y)$.
3
Substitute: $5^3 - 3(6)(5) = 125 - 90 = 35$.

Answer:

$x^3 + y^3 = 35$.

Olympiad / Challenge

Real numbers satisfy $x + y + z = 1$, $x^2 + y^2 + z^2 = 2$, and $x^3 + y^3 + z^3 = 3$. Find $x^4 + y^4 + z^4$.

1
Let $e_1, e_2, e_3$ be the elementary symmetric sums and $p_k$ the power sums. Then $e_1 = p_1 = 1$. From $p_2 = e_1 p_1 - 2e_2$: $2 = 1 - 2e_2$, so $e_2 = -\tfrac{1}{2}$.
2
From $p_3 = e_1 p_2 - e_2 p_1 + 3e_3$: $3 = (1)(2) - (-\tfrac12)(1) + 3e_3 = 2 + \tfrac12 + 3e_3$, so $e_3 = \tfrac{1}{6}$.
3
For three variables, $p_4 = e_1 p_3 - e_2 p_2 + e_3 p_1$ (no $e_4$ term): $p_4 = (1)(3) - (-\tfrac12)(2) + (\tfrac16)(1) = 3 + 1 + \tfrac16 = \tfrac{25}{6}$.

Answer:

$x^4 + y^4 + z^4 = \dfrac{25}{6}$.

Going Deeper

Generalization: the Fundamental Theorem of Symmetric Polynomials says EVERY symmetric polynomial is a unique polynomial in the elementary symmetric sums $e_1, e_2, \dots, e_n$; Newton's identities give the explicit recurrence converting between the $e_k$ and the power sums $p_k$.
Where it appears: AIME problems that give a few symmetric conditions and ask for another symmetric quantity; combined with Vieta it evaluates $\sum r_i^k$ for a polynomial's roots without ever finding them.
Pitfall: the conversion identities are degree-specific — $p^3 + q^3 = e_1^3 - 3e_1 e_2$ is for TWO variables; the three-variable analogue $\sum x^3 = e_1^3 - 3e_1 e_2 + 3e_3$ has an extra $+3e_3$. Using the two-variable formula on three variables silently drops that term.

Spot the Signal

Look for problems where the key step is using expressions unchanged by variable swaps to reduce variables and exploit root relationships.
You can describe the hard part as using expressions unchanged by variable swaps to reduce variables and exploit root relationships, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the expression or equation that calls for symmetric polynomials, then rewrite the givens around it.
Name the relation that makes Symmetric Polynomials legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Symmetric Polynomials just because the surface looks familiar; verify the required condition first.
Applying Symmetric Polynomials because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let symmetric polynomials reveal the algebraic structure; then compute only what remains.

Try it on:

Practice a contest problem where the key step is using expressions unchanged by variable swaps to reduce variables and exploit root relationships.