If #alpha#, #beta# and #gamma# are the roots of the equation #x^3+6x+1=0#, how do you find the polynomial whose roots are #alpha beta# , #beta gamma# and #alpha gamma#?

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Sep 24, 2017

Answer:

# x^3-6x^2-1=0.#

Explanation:

We know that, #alpha, beta, gamma# are the roots of the eqn.

#x^3+0x^2+6x+1=0.#

Then, by Vieta's Rule, we have,

#(1):alpha+beta+gamma=0; (2):alphabeta+betagamma+gammaalpha=6;(3):alphabetagamma=-1.#

Let, #alphabeta=c,betagamma=a,gammaalpha=b.#

Then, #a,b,c# are the roots of the desired poly.

Now, #(1'):a+b+c=betagamma+gammaalpha+alphabeta=6,#

#(2'):ab+bc+ca=alphabetagamma^2+alpha^2betagamma+alphabeta^2gamma,#

#=alphabetagamma(alpha+beta+gamma)=-1*0=0,# and,

#(3'): abc=(betagamma)(gammaalpha)(alphabeta)=alpha^2beta^2gamma^2=(-1)^2=1.#

From #(1'), (2'), and, (3'),# the desired poly cubic eqn. is,

#x^3-(a+b+c)x^2+(ab+bc+ca)x-abc=0, i.e., #

#x^3-6x^2+0x-1=0, or, x^3-6x^2-1=0.#

Enjoy Maths.!

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2
Sep 24, 2017

Answer:

The equation is #x^3-6x^2-1=0#

Explanation:

Let's rewrite the equation as

#x^3+0x^2+6x^1+1*x^0=0#

If #alpha#, #beta# and #gamma# are the roots of this equation, we have

#(x-alpha)(x-beta)(x-gamma)=0#

#(x^2-alphax-betax+alphabeta)(x-gamma)=0#

#x^3-(alpha+beta+gamma)x^2+(alphabeta+gammabeta+alphagamma)x-alphabetagamma=0#

Comparing this equation to the original equation,

#alpha + beta+gamma=0#

#alphabeta+gammabeta+alphagamma=6#

#alphabetagamma=-1#

The new roots are,

#alphabeta#, #betagamma# and #alphagamma#

Therefore,

#alphabeta+betagamma+alphagamma=6#

The coefficient of #x^2# is #-6#

#alphabeta*betagamma*alphagamma=(alphabetagamma)^2=(-1)^2=1#

The coefficient of #x^0# is #-1#

#alphabeta*betagamma+alphagamma*alphabeta+alphabeta*betagamma=alphabetagamma(alpha+beta+gamma)=0#

The coefficient of #x^1# is #0#

The equation is

#x^3-6x^2+0x^1-1*x^0=0#

#=>#, #x^3-6x^2-1=0#

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