Question #70432

1 Answer
Mar 24, 2017

Answer:

See below.

Explanation:

#(x+x_1)(x+x_2)(x-x_3)=x^3+(x_1+x_2+x_3)x^2+(x_1x_2+x_1 x_3+x_2x_3)x+x_1x_2x_3# so

#{(x_1+x_2+x_3=a),(x_1x_2+x_1 x_3+x_2x_3=b),(x_1x_2x_3=c):}#

We are looking for a polynomial #x^3+alpha x^2+beta x+gamma# such that

#{(x_1^3+x_2^3+x_3^3=alpha),(x_1^3x_2^3+x_1^3 x_3^3+x_2^3x_3^3=beta),(x_1^3x_2^3x_3^3=gamma):}#

Clearly we have

#gamma = c^3#

now #(x_1+x_2+x_3)^3 = x_1^3+x_2^3+x_3^3+3(x_1^2(x_2+x_3)+x_2^2(x_1+x_3)+x_3^2(x_1+x_2))+6 x_1x_2x_3#

but

#x_1^2(x_2+x_3)+x_2^2(x_1+x_3)+x_3^2(x_1+x_2)=(x_1+x_2+x_3)(x_1x_2+x_1 x_3+x_2x_3)+3x_1x_2x_3#

so we have

#a^3=x_1^3+x_2^3+x_3^3+3(ab+3c)+6c#

and then

#alpha = a^3-3(ab+3c)-6c#

now making

#(x_1x_2+x_1 x_3+x_2x_3)^3=x_1^3x_2^3+x_1^3 x_3^3+x_2^3x_3^3+3 (x_1^3 x_2^2 x_3 + x_1^2 x_2^3 x_3 + x_1^3 x_2 x_3^2 + x_1 x_2^3 x_3^2 + x_1^2 x_2 x_3^3 + x_1 x_2^2 x_3^3)+6x_1^2x_2^2x_3^2#

but

#x_1^3 x_2^2 x_3 + x_1^2 x_2^3 x_3 + x_1^3 x_2 x_3^2 + x_1 x_2^3 x_3^2 + x_1^2 x_2 x_3^3 + x_1 x_2^2 x_3^3=x_1x_2x_2(x_1^2(x_2+x_3)+x_2^2(x_1+x_3)+x_3^2(x_1+x_2))#

so finally

#b^3=beta+9c(ab+3c)+6c^3#

and then

#beta = b^3-9c(ab+3c)-6c^3# so the sought polynomial is

#x^3+(a^3-3(ab+3c)-6c)x^2+(b^3-9c(ab+3c)-6c^3)x+c^3#