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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#

1 Answer
Mar 13, 2017

Answer:

#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