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Module 2 · L8 of 15 ~35 min ⚡ +95 XP available

Further Sums and Products — Symmetric Identities

Vieta's formulas give you $\alpha+\beta$, $\alpha\beta$ — but what about $\alpha^3+\beta^3$, or $1/\alpha + 1/\beta$, or a brand new equation whose roots are $\alpha^2$ and $\beta^2$? Symmetric identities are algebraic bridges: they express any symmetric function of the roots entirely in terms of the Vieta values you already know, without ever solving the original equation.

Today's hook — If $\alpha$ and $\beta$ are roots of $x^2 - 5x + 6 = 0$, can you find $\alpha^3 + \beta^3$ without solving for $\alpha$ and $\beta$? Most students reach for the quadratic formula. But there's a one-line identity that bypasses all of that. By the end of this lesson you'll have a toolkit of identities that handle any power — and let you build brand new equations from transformed roots.

0/5QUESTS

Recall — your gut answer first

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$\alpha$ and $\beta$ are roots of $x^2 - 5x + 6 = 0$. Without solving for $\alpha$ and $\beta$, try to find $\alpha^3 + \beta^3$. Can you express this using only $\alpha+\beta$ and $\alpha\beta$?

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The two moves

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Every symmetric identity problem uses two moves: first write down the Vieta values ($S = \alpha+\beta$, $P = \alpha\beta$), then substitute into the correct identity to find the expression asked.

The key insight is that every symmetric polynomial in $\alpha$ and $\beta$ can be written purely in terms of $S = \alpha+\beta$ and $P = \alpha\beta$. This means you never need the individual roots — you just need two numbers from Vieta's formulas and the right identity. For transformed-roots questions, a third step appears: compute the new $S'$ and $P'$, then write the new equation.

$\alpha^3+\beta^3 = S^3 - 3PS$

Power identity shorthand

Let $S = \alpha+\beta$ and $P = \alpha\beta$. Then $\alpha^2+\beta^2 = S^2-2P$ and $\alpha^3+\beta^3 = S^3-3PS$. Write these as one-line formulas.

Reciprocal identity

$\dfrac{1}{\alpha}+\dfrac{1}{\beta} = \dfrac{\alpha+\beta}{\alpha\beta} = \dfrac{S}{P}$. New equation with roots $1/\alpha, 1/\beta$: multiply original through by $x^n$ then substitute $x \to 1/x$.

Transformed roots

To form an equation with roots $\alpha^2, \beta^2$: new sum $= \alpha^2+\beta^2 = S^2-2P$; new product $= \alpha^2\beta^2 = P^2$. Write $x^2 - (\text{new sum})\,x + (\text{new product}) = 0$.

What you'll master

Know

Key facts

$\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$
$\alpha^3+\beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)$
$\frac{1}{\alpha}+\frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta}$ and $(\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta$

Understand

Concepts

How higher-power identities are derived from lower-power ones
Why every symmetric expression can be written using only $S$ and $P$
How forming equations with transformed roots uses the same ideas

Can do

Skills

Evaluate $\alpha^2+\beta^2$, $\alpha^3+\beta^3$, $1/\alpha+1/\beta$ from Vieta values
Form a quadratic equation whose roots are $\alpha^2, \beta^2$ or $1/\alpha, 1/\beta$
Apply cubic symmetric identities to find $\alpha^2+\beta^2+\gamma^2$

Key terms

Symmetric identityAn algebraic identity whose value is unchanged when any two roots are swapped — e.g. $\alpha^2+\beta^2$ is symmetric because swapping $\alpha$ and $\beta$ gives the same result.

Elementary symmetric polynomials$S = \alpha+\beta$ and $P = \alpha\beta$ for a quadratic. Every symmetric function of $\alpha, \beta$ can be expressed in terms of $S$ and $P$.

Transformed rootsNew roots formed by applying an operation to the original roots — e.g. $\alpha^2$ and $\beta^2$, or $1/\alpha$ and $1/\beta$.

Power sum identityAn identity expressing $\alpha^n+\beta^n$ in terms of $S$ and $P$. The key ones are $n=2$ and $n=3$.

Discriminant$(\alpha-\beta)^2 = S^2 - 4P = b^2 - 4ac$. A measure of whether the roots are real and distinct, equal, or complex.

Newton's identitiesA general system for expressing power sums $\alpha^n+\beta^n+\cdots$ in terms of elementary symmetric polynomials. The identities in this lesson are the first few cases.

Key symmetric identities — quadratics

core concept

Let $S = \alpha+\beta$ and $P = \alpha\beta$ (the Vieta values). The following identities hold for any two numbers $\alpha, \beta$:

$$\alpha^2 + \beta^2 = S^2 - 2P$$

$$\alpha^3 + \beta^3 = S^3 - 3PS \quad \text{(or: } (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)\text{)}$$

$$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{S}{P} \qquad (\text{requires } P \ne 0)$$

$$\alpha^2\beta + \alpha\beta^2 = PS$$

$$(\alpha - \beta)^2 = S^2 - 4P$$

How to derive $\alpha^2+\beta^2 = S^2-2P$: Expand $(\alpha+\beta)^2 = \alpha^2+2\alpha\beta+\beta^2$. Rearranging: $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = S^2-2P$.

How to derive $\alpha^3+\beta^3 = S^3-3PS$: Expand $(\alpha+\beta)^3 = \alpha^3+3\alpha^2\beta+3\alpha\beta^2+\beta^3 = \alpha^3+\beta^3+3\alpha\beta(\alpha+\beta)$. So $\alpha^3+\beta^3 = S^3-3PS$.

Why symmetric identities matter in competition and HSC. Many exam questions ask for expressions like $\alpha^3+\beta^3$ where finding $\alpha$ and $\beta$ individually would be messy (irrational or complex). Symmetric identities let you sidestep all of that and get a clean numerical answer in seconds.

Let S = + and P = . Then:; ^2+^2 = S^2-2P

Pause — copy the quadratic symmetric identities into your book: $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = S^2-2P$; $\alpha^3+\beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta) = S^3 - 3PS$.

Quick check: If $\alpha+\beta = 5$ and $\alpha\beta = 3$, what is $\alpha^2+\beta^2$?

Key symmetric identities — cubics

core concept

We just saw that $\alpha^2 + \beta^2 = S^2 - 2P$ for a quadratic. That raises a question: for a cubic with three roots, the analogous identity for $\alpha^2 + \beta^2 + \gamma^2$ involves $S_1 = \alpha+\beta+\gamma$ and $S_2 = \alpha\beta+\beta\gamma+\gamma\alpha$ — what is the identity and how is it derived? This card answers it → $\alpha^2 + \beta^2 + \gamma^2 = S_1^2 - 2S_2$; expand $(\alpha+\beta+\gamma)^2$ and subtract the twice-the-pairwise-sum terms.

For three roots $\alpha, \beta, \gamma$ with $S_1 = \alpha+\beta+\gamma$, $S_2 = \alpha\beta+\beta\gamma+\gamma\alpha$ and $P = \alpha\beta\gamma$:

$$\alpha^2+\beta^2+\gamma^2 = S_1^2 - 2S_2$$

$$\alpha^3+\beta^3+\gamma^3 - 3P = S_1(\alpha^2+\beta^2+\gamma^2 - S_2)$$

The first identity follows by the same method as for quadratics: expand $(\alpha+\beta+\gamma)^2 = \alpha^2+\beta^2+\gamma^2 + 2(\alpha\beta+\beta\gamma+\gamma\alpha)$, then rearrange. This identity was used in Lesson 7.

For the $\alpha^3+\beta^3+\gamma^3$ identity, note that once you have $\alpha^2+\beta^2+\gamma^2$ from the first formula, you can substitute. These identities are part of Newton's power-sum recurrence — each power sum is expressible from the ones below it.

Only memorise the ones you'll use. The HSC regularly tests $\alpha^2+\beta^2$ and $\alpha^3+\beta^3$ for quadratics, and $\alpha^2+\beta^2+\gamma^2$ for cubics. The reciprocal identity $1/\alpha+1/\beta = S/P$ also appears. The cubic $\alpha^3+\beta^3+\gamma^3$ formula rarely appears at Year 11; focus on the first two cubic identities.

For cubics: ^2+^2+^2 = (++)^2 - 2(++) = S_1^2 - 2S_2; Derivation: expand (S_1)^2 and subtract twice the pairwise sum.

Pause — copy the cubic symmetric identity into your book: $\alpha^2+\beta^2+\gamma^2 = S_1^2 - 2S_2$ where $S_1 = \alpha+\beta+\gamma$ and $S_2 = \alpha\beta+\beta\gamma+\gamma\alpha$; derivation — expand $S_1^2$ and subtract $2S_2$.

Did you get this? True or false: $\alpha^3 + \beta^3 = (\alpha+\beta)^3$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · POWER SUM

If $\alpha, \beta$ are roots of $x^2 - 4x + 1 = 0$, find $\alpha^3 + \beta^3$.

$S = \alpha+\beta = 4$, $P = \alpha\beta = 1$

Read directly from the monic quadratic: $-b = 4$, $c = 1$.

PROBLEM 2 · TRANSFORMED ROOTS

If $\alpha, \beta$ are roots of $x^2 - 4x + 1 = 0$, form the equation with roots $\alpha^2$ and $\beta^2$.

$S = 4$, $P = 1$. New sum: $\alpha^2+\beta^2 = S^2-2P = 16-2 = 14$

The new sum of roots equals the identity for $\alpha^2+\beta^2$.

PROBLEM 3 · RECIPROCAL ROOTS

Form the equation with roots $\dfrac{1}{\alpha}$ and $\dfrac{1}{\beta}$ where $\alpha, \beta$ are roots of $2x^2 - 5x + 1 = 0$.

$S = \alpha+\beta = \dfrac{5}{2}$, $P = \alpha\beta = \dfrac{1}{2}$

Vieta's: $S = -b/a = 5/2$, $P = c/a = 1/2$.

Fill the gap: If $\alpha+\beta = 3$ and $\alpha\beta = 2$, then $\alpha^3 + \beta^3 = $ .

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Writing $\alpha^3+\beta^3 = (\alpha+\beta)^3$

The cube of a sum includes a middle term: $(\alpha+\beta)^3 = \alpha^3+3\alpha^2\beta+3\alpha\beta^2+\beta^3$. So $\alpha^3+\beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta) = S^3 - 3PS$. Never drop the $3\alpha\beta(\alpha+\beta)$ term.

Trap 02

Forming the wrong equation for transformed roots

For roots $\alpha^2, \beta^2$, students write the product as $(\alpha^2)(\beta^2) = \alpha^2+\beta^2$ (confusing sum and product). The product is $\alpha^2\beta^2 = (\alpha\beta)^2 = P^2$ — not the sum. Keep track of which is which by labelling them explicitly.

Trap 03

Applying the reciprocal identity when $P = 0$

$1/\alpha + 1/\beta = S/P$ requires $P = \alpha\beta \ne 0$ — i.e. neither root is zero. If $\alpha\beta = 0$ then one root is zero and $1/\alpha$ or $1/\beta$ is undefined. Always check $P \ne 0$ before using this identity.

Did you get this? True or false: if $\alpha, \beta$ are roots of $x^2 - 3x + 2 = 0$, then the equation with roots $\alpha^2, \beta^2$ is $x^2 - 5x + 4 = 0$.

Odd one out: Three of these are valid symmetric identities for roots $\alpha, \beta$. Which one is incorrect (the odd one out)?

Work mode · how are you completing this lesson?

Activities · practice with the ideas

If $\alpha, \beta$ are roots of $x^2 - 3x + 2 = 0$, find $\alpha^2+\beta^2$.

If $\alpha, \beta$ are roots of $x^2 - 3x + 2 = 0$, find $\alpha^3+\beta^3$.

Form the equation with roots $\alpha^2, \beta^2$ where $\alpha, \beta$ are roots of $x^2 - 3x + 2 = 0$.

If $\alpha, \beta, \gamma$ are roots of $x^3 - 2x^2 + x - 3 = 0$, find $\alpha^2+\beta^2+\gamma^2$.

Explain why we can always express any symmetric function of the roots purely in terms of the coefficients of the polynomial.

Revisit your thinking

Earlier you were asked for $\alpha^3+\beta^3$ when $\alpha, \beta$ are roots of $x^2-5x+6=0$. The answer: $S=5$, $P=6$, so $\alpha^3+\beta^3 = 5^3 - 3(6)(5) = 125-90 = 35$. (The roots are 2 and 3, so $\alpha^3+\beta^3 = 8+27 = 35$ — confirmed.) Every symmetric function of the roots can be written in terms of the coefficients because the coefficients are themselves the elementary symmetric polynomials of the roots — that is Vieta's theorem in its deepest form.

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Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 42 marks

Q1. If $\alpha, \beta$ are roots of $x^2 - 3x + 2 = 0$, find $\alpha^3 + \beta^3$. (2 marks)

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ApplyBand 53 marks

Q2. Form the equation with roots $\dfrac{1}{\alpha}$ and $\dfrac{1}{\beta}$ where $\alpha, \beta$ are roots of $2x^2 - 5x + 1 = 0$. (3 marks)

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ApplyBand 42 marks

Q3. If $\alpha, \beta, \gamma$ are roots of $x^3 - 2x^2 + x - 3 = 0$, find $\alpha^2 + \beta^2 + \gamma^2$. (2 marks)

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Comprehensive answers (click to reveal)

Activity answers: 1. $\alpha^2+\beta^2 = 9-4 = 5$. · 2. $\alpha^3+\beta^3 = 27-18 = 9$. · 3. New sum $= 5$, new product $= 4$; equation $x^2-5x+4=0$. (Roots are 1 and 4, i.e. $1^2$ and $2^2$.) · 4. $S_1=2$, $S_2=1$; $\alpha^2+\beta^2+\gamma^2 = 4-2 = 2$. · 5. Because every symmetric polynomial can be written in terms of the elementary symmetric polynomials $S$ and $P$ (or $S_1, S_2, P$ for cubics), and Vieta's formulas express these directly as ratios of coefficients.

Q1 (2 marks): $S=3$, $P=2$ [0.5]. $\alpha^3+\beta^3 = 27-18 = 9$ [1.5].

Q2 (3 marks): $S = 5/2$, $P = 1/2$ [0.5]. New sum $= (5/2)/(1/2) = 5$ [1]. New product $= 1/(1/2) = 2$ [0.5]. Equation: $x^2-5x+2=0$ [1].

Q3 (2 marks): $S_1 = 2$, $S_2 = 1$ [0.5]. $\alpha^2+\beta^2+\gamma^2 = 4-2 = 2$ [1.5].

Boss battle · The Identity Weaver

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Five timed questions on symmetric identities and transformed roots. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering symmetric identity questions. Lighter alternative to the boss.

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