Question

Find the cubic equation whose roots are the cubes of the roots of `x^3+ax^2+bx+c=0`,`a,b,cinRR`?

Find the cubic equation whose roots are the cubes of the roots of
`x^3+ax^2+bx+c=0`,`a,b,cinRR`

Answer 1

`x^3+(a^3-3ab+3c)x^2+(b^3-3abc+3c^2)x+c^3=0`

Explanation:

Suppose the roots of the original cubic are `alpha`, `beta` and `gamma`.

Then:

`x^3+ax^2+bx+c = (x-alpha)(x-beta)(x-gamma)`

`color(white)(x^3+ax^2+bx+c) = x^3-(alpha+beta+gamma)x^2+(alphabeta+betagamma+gammaalpha)x-alphabetagamma`

So we have:

`{ (alpha+beta+gamma = -a), (alphabeta+betagamma+gammaalpha = b), (alphabetagamma = -c) :}`

The cubic we are looking for is:

`(x-alpha^3)(x-beta^3)(x-gamma^3)`

`= x^3-(alpha^3+beta^3+gamma^3)x^2+(alpha^3beta^3+beta^3gamma^3+gamma^3alpha^3)x-alpha^3beta^3gamma^3`

So the problem essentially boils down to expressing each of the symmetric polynomials in `alpha^3`, `beta^3` and `gamma^3` constituting these coefficients in terms of the elementary symmetric polynomials in `alpha`, `beta` and `gamma`.

For example:

`(alpha+beta+gamma)^3`

`=alpha^3+beta^3+gamma^3+3(alpha^2beta+beta^2gamma+gamma^2alpha+alphabeta^2+betagamma^2+gammaalpha^2)+6alphabetagamma`

`(alpha+beta+gamma)(alphabeta+betagamma+gammaalpha)`

`=alpha^2beta+beta^2gamma+gamma^2alpha+alphabeta^2+betagamma^2+gammaalpha^2+3alphabetagamma`

So:

`alpha^3+beta^3+gamma^3`

`=(alpha+beta+gamma)^3-3(alpha+beta+gamma)(alphabeta+betagamma+gammaalpha)+3alphabetagamma`

`=-a^3+3ab-3c`

We also find:

`(alphabeta+betagamma+gammaalpha)^3`

`= alpha^3beta^3+beta^3gamma^3+gamma^3alpha^3+3(alpha^3beta^2gamma+beta^3gamma^2alpha+gamma^3alpha^2beta+alpha^3betagamma^2+beta^3gammaalpha^2+gamma^3alphabeta^2)+6alpha^2beta^2gamma^2`

`(alpha+beta+gamma)(alphabeta+betagamma+gammaalpha)alphabetagamma`

`=alpha^3beta^2gamma+beta^3gamma^2alpha+gamma^3alpha^2beta+alpha^3betagamma^2+beta^3gammaalpha^2+gamma^3alphabeta^2+3alpha^2beta^2gamma^2`

So:

`alpha^3beta^3+beta^3gamma^3+gamma^3alpha^3`

`=(alphabeta+betagamma+gammaalpha)^3-3(alpha+beta+gamma)(alphabeta+betagamma+gammaalpha)alphabetagamma+3(alphabetagamma)^2`

`=b^3-3abc+3c^2`

Finally:

`alpha^3beta^3gamma^3 = (alphabetagamma)^3 = -c^3`

Hence the required cubic equation is:

`x^3+(a^3-3ab+3c)x^2+(b^3-3abc+3c^2)x+c^3 = 0`

`color(white)()`
Footnote

If you would like to see a more advanced application of symmetric polynomials, you may like to take a look at this one: https://socratic.org/s/aCWXbG2b