Algebra theory - symmetric expressions example questions

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Example 1:

$x^3 + \rho x + q = 0$
$\alpha, \beta, \gamma$ are roots of the cubic.

Prove that

$$\frac{\alpha^5 + \beta^5 + \gamma^5}{5} = \frac{\alpha^3 + \beta^3 + \gamma^3}{3} \cdot \frac{\alpha^2 + \beta^2 + \gamma^2}{2}$$

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$$x^3 + 0x^2 + \rho x + q = 0$$ $$\alpha + \beta + \gamma = 0 \Rightarrow (\alpha + \beta + \gamma)^2 = 0 \Rightarrow \alpha^2 + \beta^2 + \gamma^2 + 2(\alpha\beta + \beta\gamma + \gamma\alpha) = 0 \Rightarrow \alpha^2 + \beta^2 + \gamma^2 = -2(\alpha\beta + \beta\gamma + \gamma\alpha)$$ $$\alpha\beta + \beta\gamma + \gamma\alpha = \rho \Rightarrow \alpha^2 + \beta^2 + \gamma^2 = -2\rho$$

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$$\alpha^3 + \rho \alpha + q = 0 \Rightarrow \alpha^3 = -\rho\alpha - q,\quad \beta^3 = -\rho\beta - q,\quad \gamma^3 = -\rho\gamma - q$$ $$\alpha^3 + \beta^3 + \gamma^3 = -\rho(\alpha + \beta + \gamma) - 3q \Rightarrow \alpha^3 + \beta^3 + \gamma^3 = -3q$$ $$\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) = 0 - 2\rho \Rightarrow \alpha^2 + \beta^2 + \gamma^2 = -2\rho$$

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$$\alpha^3 + \rho \alpha + q = 0 \Rightarrow \alpha^2(\alpha^3 + \rho \alpha + q) = 0 \cdot \alpha^2 \Rightarrow \alpha^5 + \rho \alpha^3 + q\alpha^2 = 0 \Rightarrow \alpha^5 = -\rho \alpha^3 - q\alpha^2$$

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$$\alpha^5 = -\rho \alpha^3 - q\alpha^2$$ $$= -\rho(-\rho \alpha - q) - q \alpha^2 = \rho^2 \alpha + \rho q - q \alpha^2 \Rightarrow \rho^2 \alpha + \rho q - q \alpha^2$$ $$\alpha^5 = -q \alpha^2 + \rho^2 \alpha + \rho q$$ $$\beta^5 = -q \beta^2 + \rho^2 \beta + \rho q$$ $$\gamma^5 = -q \gamma^2 + \rho^2 \gamma + \rho q$$

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$$\alpha^5 + \beta^5 + \gamma^5 = -q(\alpha^2 + \beta^2 + \gamma^2) + \rho^2(\alpha + \beta + \gamma) + 3\rho q$$ $$= -q(-2\rho) + 0 + 3\rho q$$ $$\alpha^5 + \beta^5 + \gamma^5 = 5\rho q$$

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$$\frac{\alpha^5 + \beta^5 + \gamma^5}{5} = (-\rho)(-q) = \frac{\alpha^3 + \beta^3 + \gamma^3}{3} \cdot \frac{\alpha^2 + \beta^2 + \gamma^2}{2}$$

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Example 2:

$$\alpha + \beta + \gamma = 0 \quad \text{then}$$ $$\text{Find value of } \frac{3(\alpha^2 + \beta^2 + \gamma^2)(\alpha^5 + \beta^5 + \gamma^5)}{(\alpha^3 + \beta^3 + \gamma^3)(\alpha^4 + \beta^4 + \gamma^4)}$$

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Solution:
Assume a cubic polynomial whose roots are 

$\alpha, \beta, \gamma$:

$$a x^3 + b x^2 + c x + d = 0$$

Given $\alpha + \beta + \gamma = 0 \Rightarrow \frac{b}{a} = 0 \Rightarrow b = 0$

$$a x^3 + c x + d = 0 \Rightarrow x^3 + \frac{c}{a} x + \frac{d}{a} = 0 \Rightarrow x^3 + \rho x + q = 0$$ $$\alpha + \beta + \gamma = 0 \quad \alpha\beta + \beta\gamma + \gamma\alpha = \rho \quad \alpha\beta\gamma = -q$$ $$\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) = 0 - 2(\rho)$$ $$\Rightarrow \alpha^2 + \beta^2 + \gamma^2 = -2\rho$$ $$\alpha^3 + \beta^3 + \gamma^3 = -3q$$ $$\alpha^5 + \beta^5 + \gamma^5 = 5\rho q$$

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$$\alpha^3 + \rho \alpha + q = 0$$ $$\alpha^4 + \rho \alpha^2 + q \alpha = 0 \Rightarrow \alpha^4 = -\rho \alpha^2 - q \alpha$$ $$\beta^4 = -\rho \beta^2 - q \beta \quad \gamma^4 = -\rho \gamma^2 - q \gamma$$

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$$\alpha^4 + \beta^4 + \gamma^4 = -\rho(\alpha^2 + \beta^2 + \gamma^2) - q(\alpha + \beta + \gamma) = \rho(-2\rho) - q(0)$$ $$\Rightarrow \alpha^4 + \beta^4 + \gamma^4 = 2\rho^2$$

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$$\frac{3(-2\rho)(5\rho q)}{(-3q)(2\rho^2)} = 5$$

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Example 3:

Problem Statement:

Given $x, y, z$ are real and:

$$x + y + z = 3$$ $$x^2 + y^2 + z^2 = 5$$ $$x^3 + y^3 + z^3 = 7$$

Find the value of $x^4 + y^4 + z^4$

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Solution:

Given:

$$x + y + z = 3$$ $$xy + yz + zx = 2 \quad \text{(Derived)}$$ $$xyz = -\frac{2}{3}$$

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From the identity:

$$(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$$ $$9 = 5 + 2(xy + yz + zx) \Rightarrow xy + yz + zx = 2$$

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Using identity:

$$x^3 + y^3 + z^3 - 3xyz = (x + y + z)[x^2 + y^2 + z^2 - (xy + yz + zx)]$$ $$7 - 3xyz = 3[5 - 2] \Rightarrow 7 - 3xyz = 9 \Rightarrow xyz = -\frac{2}{3}$$

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Starting from the cubic:

$$x^3 - 3x^2 + 2x + \frac{2}{3} = 0$$

This polynomial holds for $x, y, z$, so:

$$x^3 - 3x^2 + 2x + \frac{2}{3} = 0 \Rightarrow x^4 = 3x^3 - 2x^2 - \frac{2}{3}x$$

Similarly:

$$y^4 = 3y^3 - 2y^2 - \frac{2}{3}y$$

$z^4$ follows same form.

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Now compute:

$$x^4 + y^4 + z^4 = 3(x^3 + y^3 + z^3) - 2(x^2 + y^2 + z^2) - \frac{2}{3}(x + y + z)$$

Substituting known values:

$$= 3(7) - 2(5) - \frac{2}{3}(3) = 21 - 10 - 2 = 9$$

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