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
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.

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 }