Equation problem

  • Thread starter DoremiCSD
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  • #1
DoremiCSD
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Homework Statement


If the equation 2x ^ 3 - 15X ^ 2 + 30 x - 7 = 0 has roots a, b ​​and c to find the value of the expression a ^ 3 + b ^ 3 + c ^ 3. Also to be found without using the types Vieta,the equation with roots 1 / (a-3), 1 / (b-3), 1 / (c-3)


Homework Equations


I found the first part of a^3+b^3+c^3=754/8 but i cant find the second part of the exercise without using the type of vieta any ideas?


The Attempt at a Solution

 

Answers and Replies

  • #2
Bacle2
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Wouldn't the equation be given by:

(x-r1)(x-r2)(x-r3) ?
 
  • #3
DoremiCSD
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yea but he (x-r1)(x-r2)(x-r3) conclude to the types of vietta.. he says to find the equation without to use vietta
 
  • #4
NascentOxygen
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Homework Statement


If the equation 2x ^ 3 - 15X ^ 2 + 30 x - 7 = 0 has roots a, b ​​and c to find the value of the expression a ^ 3 + b ^ 3 + c ^ 3.
What method did you use to determine this?
 
  • #5
DoremiCSD
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i used the types of vietta and i found abc,a+b+c,ab+bc+ca cuz its says to not use vietta types only in second statement and then i used eulers equation

a^3+b^3+c^3=3abc+(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
 
  • #6
NascentOxygen
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i used the types of vietta and i found abc,a+b+c,ab+bc+ca cuz its says to not use vietta types only in second statement and then i used eulers equation

a^3+b^3+c^3=3abc+(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
So you as good as solved for the individual roots, then multiplied all 3 to get abc, and added all 3 to get a+b+c, etc?
 
  • #7
DoremiCSD
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yea as you know vietta is formed like this

abc=-d/a, a+b+c=-b/a, ab+bc+ca=c/a
 
  • #8
NascentOxygen
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So as you have determined 'a', can you now subtract 3 from that value of 'a' and take the reciprocal to get the new root r1, same for 'b' and 'c', and then multiply out (x-r1)(x-r2)(x-r3) ?
 
  • #9
DoremiCSD
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yea i understand but maybe this (x-r1)(x-r2)(x-r3) guide to vietta types? or not?
 
  • #10
NascentOxygen
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Is it permissible to work it out from basics, making no reliance on Mr Vieté's formulae? :smile:

Multiplying out (x-(a-3)⁻¹)·(x-(b-3)⁻¹)·(x-(c-3)⁻¹), collecting terms and setting the coefficient of x³ to unity, I can see that the coefficient of x² here can be obtained from the earlier equation as its coefficient of x plus 6 times its coefficient of x² plus a constant. The coefficient of x here can similarly be readily seen.

The difference from that earlier is that there is now no need to have actually solved for a,b, and c.
 

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