Solution for the problem Watch

Shas72
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pls help me with this sum.
The cubic equation x^3-x^2+3=0 has roots alpha, beta and gamma.
By considering s1 and s4 determine the value of alpha ^3(beta+alpha)+beta^3(alpha+gamm a)+gamma^3(alpha +beta).
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Shas72
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pls help me with this sum.
The cubic equation x^3-x^2+3=0 has roots alpha, beta and gamma.
By considering s1 and s4 determine the value of alpha ^3(beta+alpha)+beta^3(alpha+gamm a)+gamma^3(alpha +beta).
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RDKGames
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(Original post by Shas72)
pls help me with this sum.
The cubic equation x^3-x^2+3=0 has roots alpha, beta and gamma.
By considering s1 and s4 determine the value of alpha ^3(beta+alpha)+beta^3(alpha+gamm a)+gamma^3(alpha +beta).
What are s1 and s4 defined to be?
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Shas72
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(Original post by RDKGames)
What are s1 and s4 defined to be?
The first question in that is using a relation Sn=alpha^n+beta^n+gamma^n find the value of s4.
I found this value of s4 and it is 2.
The next question that I have posted is what iam not able to solve
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Shas72
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(Original post by Shas72)
The first question in that is using a relation Sn=alpha^n+beta^n+gamma^n find the value of s4.
I found this value of s4 and it is 2.
The next question that I have posted is what iam not able to solve
S1=-b/a which is 0
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RDKGames
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(Original post by Shas72)
The first question in that is using a relation Sn=alpha^n+beta^n+gamma^n find the value of s4.
I found this value of s4 and it is 2.
The next question that I have posted is what iam not able to solve
Okay, so I trust you got s_4 = \alpha^4 + \beta^4 + \gamma^4 = 2 correct. We also have s_1 = \alpha + \beta + \gamma = -1.

Also, I think in your question you made a typo and instead want to find

\alpha^3(\beta+\gamma)+\beta^3( \alpha +\gamma)+\gamma^3(\alpha +\beta), right?
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Shas72
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(Original post by RDKGames)
Okay, so I trust you got s_4 = \alpha^4 + \beta^4 + \gamma^4 = 2 correct. We also have s_1 = \alpha + \beta + \gamma = -1.

Also, I think in your question you made a typo and instead want to find

\alpha^3(\beta+\gamma)+\beta^3( \alpha +\gamma)+\gamma^3(\alpha +\beta), right?
Yeah you are right. Iam sorry for typo error
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Shas72
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(Original post by RDKGames)
Okay, so I trust you got s_4 = \alpha^4 + \beta^4 + \gamma^4 = 2 correct. We also have s_1 = \alpha + \beta + \gamma = -1.

Also, I think in your question you made a typo and instead want to find

\alpha^3(\beta+\gamma)+\beta^3( \alpha +\gamma)+\gamma^3(\alpha +\beta), right?
And s1 is 0 right as there is no x^2
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RDKGames
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(Original post by Shas72)
And s1 is 0 right as there is no x^2
Yes there is, unless that's another one of your typos.
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Shas72
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The sum is x^3-x+3=0 has roots alpha beta and gamma.

So sum of roots s1 will be -b/a which is 0 as there is no x^2 .Right?
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RDKGames
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(Original post by Shas72)
The sum is x^3-x+3=0 has roots alpha beta and gamma.

So sum of roots s1 will be -b/a which is 0 as there is no x^2 .Right?
In that case, yes. In your original posts you said x^3-x^2+3=0 instead.

Anyway, so we have s_1 = 0 and s_4 = 2 which you got correct, hopefully.

Then for the rest of the question, you should consider the product

(\alpha^3 + \beta^3 + \gamma^3)(\alpha + \beta + \gamma)

and expand it appropriately so that you end up with the wanted result somewhere in the expansion.
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Shas72
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Yeah I did that and iam stuck in the last step that you have shown where you are telling me to do the expansion.
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RDKGames
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(Original post by Shas72)
Yeah I did that and iam stuck in the last step that you have shown where you are telling me to do the expansion.
(\alpha^3 + \beta^3 + \gamma^3)(\alpha + \beta + \gamma) = \alpha^4 + \alpha^3(\beta + \gamma) + \beta^4 + \beta^3(\alpha + \gamma) + \gamma^4 + \gamma^3(\alpha + \beta)

Proceed from there.
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Shas72
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But when you do that you will get 2+0 is 2.but the answer is -2.
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RDKGames
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(Original post by Shas72)
But when you do that you will get 2+0 is 2.but the answer is -2.
Not sure what you mean...

Rearranging the above we have

(\alpha^3 + \beta^3 + \gamma^3)\underbrace{(\alpha + \beta + \gamma)}_{=0} = \underbrace{(\alpha^4 + \beta^4 + \gamma^4)}_{=2} + \alpha^3(\beta + \gamma)  + \beta^3(\alpha + \gamma)  + \gamma^3(\alpha + \beta)

So the answer *is* -2.
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Shas72
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Oh wow!!! That's how we do that. Thank you so much. Thanks a tonne
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Shas72
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(Original post by RDKGames)
Not sure what you mean...

Rearranging the above we have

(\alpha^3 + \beta^3 + \gamma^3)\underbrace{(\alpha + \beta + \gamma)}_{=0} = \underbrace{(\alpha^4 + \beta^4 + \gamma^4)}_{=2} + \alpha^3(\beta + \gamma)  + \beta^3(\alpha + \gamma)  + \gamma^3(\alpha + \beta)

So the answer *is* -2.
Thanks again and again!
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