# Solution for the problemWatch

#1
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).
0
#2
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).
0
4 weeks ago
#3
(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?
0
#4
(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
0
#5
(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
0
4 weeks ago
#6
(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 correct. We also have .

, right?
0
#7
(Original post by RDKGames)
Okay, so I trust you got correct. We also have .

, right?
Yeah you are right. Iam sorry for typo error
0
#8
(Original post by RDKGames)
Okay, so I trust you got correct. We also have .

, right?
And s1 is 0 right as there is no x^2
0
4 weeks ago
#9
(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.
0
#10
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?
0
4 weeks ago
#11
(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 and which you got correct, hopefully.

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

and expand it appropriately so that you end up with the wanted result somewhere in the expansion.
0
#12
Yeah I did that and iam stuck in the last step that you have shown where you are telling me to do the expansion.
0
4 weeks ago
#13
(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.

Proceed from there.
0
#14
But when you do that you will get 2+0 is 2.but the answer is -2.
0
4 weeks ago
#15
(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

0
#16
Oh wow!!! That's how we do that. Thank you so much. Thanks a tonne
0
#17
(Original post by RDKGames)
Not sure what you mean...

Rearranging the above we have

Thanks again and again!
1
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