Solution for the problem

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).

OP

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).

Study Forum Helper

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?

OP

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

OP

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

Study Forum Helper

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?

OP

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

OP

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

Study Forum Helper

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.

OP

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?

Study Forum Helper

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.

OP

Yeah I did that and iam stuck in the last step that you have shown where you are telling me to do the expansion.

Study Forum Helper

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.

OP

But when you do that you will get 2+0 is 2.but the answer is -2.

Study Forum Helper

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.

OP

Oh wow!!! That's how we do that. Thank you so much. Thanks a tonne

OP

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!

Quick Reply

Latest

Trending

Trending

Articles for you

Your guide to handling revision and exam stress

Your guide to handling revision and exam stress

Women in Stem: interview with an apprentice engineer

Women in Stem: interview with an apprentice engineer

Why I chose a law degree and what studying law was like

Why I chose a law degree and what studying law was like

How technology is transforming education

How technology is transforming education