Solve Equation Problem w/o Vieta: Find a^3+b^3+c^3

  • Thread starter DoremiCSD
  • Start date
In summary: Instead, just multiplying out the coefficients of x² and x³ and solving for x. This is what was done in the second statement.
  • #1
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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 can't find the second part of the exercise without using the type of vieta any ideas?


The Attempt at a Solution

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

(x-r1)(x-r2)(x-r3) ?
 
  • #3
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
DoremiCSD said:

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
i used the types of vietta and i found abc,a+b+c,ab+bc+ca because 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
DoremiCSD said:
i used the types of vietta and i found abc,a+b+c,ab+bc+ca because 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
yea as you know vietta is formed like this

abc=-d/a, a+b+c=-b/a, ab+bc+ca=c/a
 
  • #8
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
yea i understand but maybe this (x-r1)(x-r2)(x-r3) guide to vietta types? or not?
 
  • #10
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.
 

1. What is the Vieta's formula?

Vieta's formula is a mathematical formula that relates the coefficients of a polynomial to the sums and products of its roots. It is used to solve polynomial equations, including finding the values of a, b, and c in the equation a^3+b^3+c^3.

2. Why is it important to solve equations without using Vieta's formula?

Solving equations without using Vieta's formula can help develop a deeper understanding of the concepts behind polynomial equations and their solutions. It also allows for alternative methods of solving equations, which can be useful in different scenarios.

3. How can I solve the equation a^3+b^3+c^3 without using Vieta's formula?

One method to solve this equation is by using the concept of perfect cubes. If we can factor the equation into the form (a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0, then we can use the quadratic formula to solve for a^2+b^2+c^2 and then find the values of a, b, and c.

4. Can I use Vieta's formula to solve this equation?

Yes, Vieta's formula can be used to solve this equation. However, as mentioned earlier, it is important to explore alternative methods to gain a better understanding of the concepts involved.

5. Are there any other methods to solve this equation?

Aside from using perfect cubes and Vieta's formula, there are other methods that can be used to solve this equation. These include using the sum and product of roots formula, substitution, and graphing. It is important to find the method that works best for you and the given equation.

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