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

[DoremiCSD]

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

 

[Bacle2]

Wouldn't the equation be given by:

(x-r1)(x-r2)(x-r3) ?

 

[DoremiCSD]

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

 

[NascentOxygen]

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?

 

[DoremiCSD]

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)

 

[NascentOxygen]

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?

 

[DoremiCSD]

yea as you know vietta is formed like this

abc=-d/a, a+b+c=-b/a, ab+bc+ca=c/a

 

[NascentOxygen]

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) ?

 

[DoremiCSD]

yea i understand but maybe this (x-r1)(x-r2)(x-r3) guide to vietta types? or not?

 

[NascentOxygen]

Is it permissible to work it out from basics, making no reliance on Mr Vieté's formulae?

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.