Help with polynomial question (1 Viewer)

qorih · Jul 12, 2015

Hello,
Having trouble working out this question:
If alpha, beta, gamma are roots of the equation x^3+2x^2-9=0
find the value of
alpha^3+beta^3+gamma^3
Any help is appreciated, thankyou

Ekman · Jul 12, 2015

qorih said:
Hello,
Having trouble working out this question:
If alpha, beta, gamma are roots of the equation x^3+2x^2-9=0
find the value of
alpha^3+beta^3+gamma^3
Any help is appreciated, thankyou

$\alpha^3 + 2\alpha^2 -9 =0 \Rightarrow \alpha^3 = 9-2\alpha^2$

$\beta^3 + 2\beta^2 -9 =0 \Rightarrow \beta^3 = 9-2\beta^2$

$\gamma^3 + 2\gamma^2 -9 =0 \Rightarrow \gamma^3 = 9-2\gamma^2$

$\therefore \alpha^3 + \beta^3 + \gamma^3 = 27-2(\alpha^2 + \beta^2 + \gamma^2)$

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

$\therefore \alpha^3 + \beta^3 + \gamma^3 = 27-2\times 4 = 19$

qorih · Jul 13, 2015

Ekman said:
$\alpha^3 + 2\alpha^2 -9 =0 \Rightarrow \alpha^3 = 9-2\alpha^2$

$\beta^3 + 2\beta^2 -9 =0 \Rightarrow \beta^3 = 9-2\beta^2$

$\gamma^3 + 2\gamma^2 -9 =0 \Rightarrow \gamma^3 = 9-2\gamma^2$

$\therefore \alpha^3 + \beta^3 + \gamma^3 = 27-2(\alpha^2 + \beta^2 + \gamma^2)$

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

$\therefore \alpha^3 + \beta^3 + \gamma^3 = 27-2\times 4 = 19$

Thankyou so much!

Help with polynomial question (1 Viewer)

qorih

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Ekman

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qorih

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