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.
$\alpha$ 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 $\alpha\beta$?
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.
Let $S = \alpha+\beta$ and $P = \alpha\beta$ (the Vieta values). The following identities hold for any two numbers $\alpha, \beta$:
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 $\alpha^3+\beta^3 = S^3-3PS$.
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 $\alpha^2+\beta^2$?
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 $P = \alpha\beta\gamma$:
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.
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
If $\alpha, \beta$ are roots of $x^2 - 4x + 1 = 0$, find $\alpha^3 + \beta^3$.
If $\alpha, \beta$ are roots of $x^2 - 4x + 1 = 0$, form the equation with roots $\alpha^2$ and $\beta^2$.
Form the equation with roots $\dfrac{1}{\alpha}$ and $\dfrac{1}{\beta}$ where $\alpha, \beta$ are roots of $2x^2 - 5x + 1 = 0$.
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
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)?
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.
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.
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. If $\alpha, \beta$ are roots of $x^2 - 3x + 2 = 0$, find $\alpha^3 + \beta^3$. (2 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)
Q3. If $\alpha, \beta, \gamma$ are roots of $x^3 - 2x^2 + x - 3 = 0$, find $\alpha^2 + \beta^2 + \gamma^2$. (2 marks)
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].
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