Having equation #x^3+alphax^2+alpha(alpha+1)+alpha^2(alpha+1)=0;alphainRR#.How to calculate #x_1^2x_2^2+x_1^2x_3^2+x_2^2x_3^2#?

2 Answers
Steve M
Jul 7, 2017

# x_1^2x_2^2 + x_1^2x_3^2 + x_2^2x_3^2 = alpha^2(1-alpha^2)#

Explanation:

If we have a cubic equation:

# ax^3+bx^2+cx+d = 0 #

With roots #alpha#, #beta# and #gamma#, then:

# alpha + beta + gamma = -b/a #
# alpha beta gamma = -d/a #
# alpha beta + beta gamma + alpha gamma = c/a #

Assuming that the correct equation is:

# x^3 + alpha x^2 + alpha(alpha+1) x + alpha^2(alpha+1) = 0 #

Assuming that #x_1#, #x_2# and #x_3# are the roots of the given cubic then we want to calculate:

# x_1^2x_2^2+x_1^2x_3^2+x_2^2x_3^2 #

Using the above properties we have:

# a = 1 #
# b = alpha #
# c = alpha(alpha+1) #
# d = alpha^2(alpha+1) #

And so the root properties are:

# x_1 + x_2 + x_3 " " = -alpha #
# x_1 x_2 x_3 " " = -alpha^2(alpha+1) #
# x_1 x_2 + x_2 x_3 + x_1 x_3 = alpha(alpha+1) #

Consider:

# (x_1x_2 + x_2x_3 + x_1x_3)^2 = x_1^2x_2^2 +x_1^2x_3^2+x_2^2x_3^2+2x_1^2x_2x_3 +2x_1x_2^2x_3 +2x_1x_2x_3^2#
# " " = x_1^2x_2^2 + x_1^2x_3^2 + x_2^2x_3^2+2x_1x_2x_3(x_1 + x_2 +x_3)#

From which it follows that:

# (alpha(alpha+1))^2 = x_1^2x_2^2 + x_1^2x_3^2 + x_2^2x_3^2 + 2(-alpha^2(alpha+1))(-alpha)#

# :. alpha^2(alpha+1)^2 = x_1^2x_2^2 + x_1^2x_3^2 + x_2^2x_3^2 + 2alpha^3(alpha+1)#

# :. x_1^2x_2^2 + x_1^2x_3^2 + x_2^2x_3^2 = alpha^2(alpha+1)^2 - 2alpha^3(alpha+1)#

# " " = alpha^2(alpha+1){(alpha+1)-2alpha}#

# " " = alpha^2(1+alpha)(1-alpha)#

# " " = alpha^2(1-alpha^2)#

Cesareo R.
Jul 7, 2017

# 2alpha^2(alpha+1)^2#

Explanation:

# (x_1 x_2)^2 + (x_1 x_3)^2 + (x_2 x_3)^2= (x_1 x_2 + x_1 x_3 + x_2 x_3)^2 -2x_1x_2x_3(x_1+x_2+x_3)#

and

#{(x_1+x_2+x_3 = -alpha),(x_1 x_2 + x_1 x_3 + x_2 x_3=0),(x_1x_2x_3 = -alpha(alpha+1)^2):}#

so

# (x_1 x_2)^2 + (x_1 x_3)^2 + (x_2 x_3)^2=-2alpha(alpha+1)^2(-alpha) = 2alpha^2(alpha+1)^2#

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