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I have a cubic expression $3x^3 - x^2 - 10x + 8$ has roots $\alpha_1, \alpha_2, \alpha_3$ and am looking to find the sum of the cubes of these roots. To do this I want to switch to a new polynomial which has roots $\alpha_1^3, \alpha_2^3, \alpha_3^3$ and then to apply the formulae for properties of roots to find their sum in terms of the coefficients of the new polynomial.

My approach was to take $x$ at one of its roots to be $u^3$ where $u$ is a root of the new polynomial, thus $x = u^{\frac{1}{3}}$. Subsituting:

$3 u - u^{\frac{2}{3}} - 10u^{\frac{1}{3}} + 8 = 0$

Now to make this a polynomial, I must change the expression so that all the exponents are integer powers. The most obvious approach would be to make a substitution $u = w^3$ so that the expression becomes $3 w^3 - w^2 - 10w + 8$. However, I cannot do this as this would give roots that are a power of 3 too low and as such I wouldn't be able to use any of the formuale to find out the sum of $α_1 + α_2+α_3$ (in fact, its just the polynomial I started with).

physBa
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  • If you are trying to find $\sum \alpha_i$, then you can just use viete's to conclude that it is $-\frac{-1}{3}=\frac{1}{3}$. Why are you trying to find this new polynomial? – Alan Abraham Aug 06 '21 at 19:07
  • Alt. hint: $;3 \sum \alpha_j^3 = \sum \alpha_j^2 + 10 \sum \alpha_j - 3 \cdot 8 = \left(\sum \alpha_j\right)^2 - 2\sum \alpha_i\alpha_j + 10 \sum \alpha_j - 24=\dots$ – dxiv Aug 06 '21 at 19:42
  • @AlanAbraham sorry there was a mistake in my post, I am trying to find the sum of the cubes of these roots. So I am trying to find a new polynomial which has the cubes as its roots and then applying viete's to this new polynomial – physBa Aug 06 '21 at 21:17
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    @physBa You don't need to find the (full) polynomial for that. See for example What is the sum of the cube of the roots of .... – dxiv Aug 06 '21 at 21:28

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