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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 confidencethat 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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