I stumbled across this... What are the values of x, y and z?

x + y + z = -1
xy + xz + yz = -2
xyz = -1

2 Answers
R. Nguyen
Jul 23, 2018

#x#, #y#, and #z# are the solutions to #w^3 + w^2 - 2w + 1 = 0#

Explanation:

Consider a cubic function of the form

#f(x) = (x - x_1)(x - x_2)(x - x_3)#.

By the fundamental theorem of algebra, we know that the three and only three roots to this function are #x_1#, #x_2#, and #x_3#.

Now, expanding the expression and grouping like terms:

#f(x) = x^3 - (color(red)(x_1 + x_2 + x_3))x^2 + (color(red)(x_1x_2 + x_1x_3 + x_2x_3))x - color(red)(x_1x_2x_3)#

Don't be intimidated - this is exactly the same polynomial, so it still has those three roots. Just because it deserves a mention, this is a specific case of Vieta's formulas .

Notice that the same expressions in the original question pop up here (but relabelled #x_1#, #x_2#, and #x-3# rather than #x#, #y#, and #z#).

#x_1 + x_2 + x_3 = -1#

#x_1x_2 + x_1x_3 + x_2x_3 = -2#

#x_1x_2x_3 = -1#

We may substitute these restrictions into our function.

#f(x) = x^3 - (color(red)(-1))x^2 + (color(red)(-2))x - (color(red)(-1))#

#f(x) = x^3 + x^2 - 2x + 1#

The roots of this function are still #x_1#, #x_2#, and #x_3#. In other words, their values are the solutions to the cubic equation

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

To answer the original question, #x#, #y#, and #z# are the solutions to #w^3 + w^2 - 2w + 1 = 0#. If you'd like, you can solve the cubic equation, but I think this answer should suffice.

George C.
Jul 23, 2018

#{ x, y, z } = { 1/3(-1+omega^n root(3)((-47+3sqrt(93))/2)+omega^-n root(3)((-47-3sqrt(93))/2)) : n = 0, 1, 2 }#

where #omega = -1/2+sqrt(3)/2i# is the primitive complex cube root of #1#.

Explanation:

To complement the other answer, here is a method for solving #w^3+w^2-2w+1=0# ...

Given:

#f(w) = w^3+w^2-2w+1#

Discriminant

The discriminant #Delta# of a cubic polynomial in the form #aw^3+bw^2+cw+d# is given by the formula:

#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#

In our example, #a=1#, #b=1#, #c=-2# and #d=1#, so we find:

#Delta = 4+32-4-27-36 = -31#

Since #Delta < 0# this cubic has #1# Real zero and #2# non-Real Complex zeros, which are Complex conjugates of one another.

Tschirnhaus transformation

To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation.

#0=27f(w)=27w^3+27w^2-54w+27#

#=(3w+1)^3-21(3w+1)+47#

#=t^3-21t+47#

where #t=(3w+1)#

Cardano's method

We want to solve:

#t^3-21t+47=0#

Let #t=u+v#.

Then:

#u^3+v^3+3(uv-7)(u+v)+47=0#

Add the constraint #v=7/u# to eliminate the #(u+v)# term and get:

#u^3+343/u^3+47=0#

Multiply through by #u^3# and rearrange slightly to get:

#(u^3)^2+47(u^3)+343=0#

Use the quadratic formula to find:

#u^3=(-47+-sqrt((47)^2-4(1)(343)))/(2*1)#

#=(-47+-sqrt(2209-1372))/2#

#=(-47+-sqrt(837))/2#

#=(-47+-3sqrt(93))/2#

Since this is Real and the derivation is symmetric in #u# and #v#, we can use one of these roots for #u^3# and the other for #v^3# to find Real root:

#t_1=root(3)((-47+3sqrt(93))/2)+root(3)((-47-3sqrt(93))/2)#

and related Complex roots:

#t_2=omega root(3)((-47+3sqrt(93))/2)+omega^2 root(3)((-47-3sqrt(93))/2)#

#t_3=omega^2 root(3)((-47+3sqrt(93))/2)+omega root(3)((-47-3sqrt(93))/2)#

where #omega=-1/2+sqrt(3)/2i# is the primitive Complex cube root of #1#.

Now #w=1/3(-1+t)#. So the roots of our original cubic are:

#w_1 = 1/3(-1+root(3)((-47+3sqrt(93))/2)+root(3)((-47-3sqrt(93))/2))#

#w_2 = 1/3(-1+omega root(3)((-47+3sqrt(93))/2)+omega^2 root(3)((-47-3sqrt(93))/2))#

#w_3 = 1/3(-1+omega^2 root(3)((-47+3sqrt(93))/2)+omega root(3)((-47-3sqrt(93))/2))#

Hence we can express the solution of the original problem as:

#{ x, y, z } = { 1/3(-1+omega^n root(3)((-47+3sqrt(93))/2)+omega^-n root(3)((-47-3sqrt(93))/2)) : n = 0, 1, 2 }#

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