How do you solve #x^3-x=1#?

2 Answers
Apr 24, 2017

Answer:

See below.

Explanation:

For my approach, I will be using a graphical interpretation.

You can rewrite the equation as #x^3 - x - 1 = 0# as the first step.

Then graph the following: #f(x) = x^3 - x - 1#.

graph{x^3 - x - 1 [-10, 10, -5, 5]}

Click on where the graph intersects the x-axis. This point should be (1.325, 0). Therefore, the answer is #x = 1.325#.

Apr 24, 2017

Answer:

Use Cardano's method to find real root:

#x_1 = 1/3(root(3)((27+3sqrt(69))/2)+root(3)((27-3sqrt(69))/2))#

#color(white)(x_1) ~~ 1.324717957#

and related complex roots.

Explanation:

Given:

#x^3-x=1#

Subtract #1# from both sides to get:

#x^3-x-1 = 0#

#color(white)()#
Discriminant

The discriminant #Delta# of a cubic polynomial in the form #ax^3+bx^2+cx+d# is given by the formula:

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

In our example, #a=1#, #b=0#, #c=-1# and #d=-1#, so we find:

#Delta = 0+4+0-27+0 = -23#

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

#color(white)()#
Cardano's method

Let #x = u+v#.

Then our equation becomes:

#(u+v)^3-(u+v)-1 = 0#

Multiplying out and rearranging a little:

#u^3+v^3+(3uv-1)(u+v)-1 = 0#

Add the constraint #v=1/(3u)# to eliminate the term in #(u+v)# and get:

#u^3+1/(27u^3)-1 = 0#

Multiply through by #27u^3# to get:

#27(u^3)^2-27(u^3)+1 = 0#

Using the quadratic formula, we find:

#u^3 = (27+-sqrt((-27)^2-4(27)(1)))/(2*27)#

#color(white)(u^3) = (27+-sqrt(729-108))/54#

#color(white)(u^3) = (27+-sqrt(621))/54#

#color(white)(u^3) = (27+-3sqrt(69))/54#

Note that these values are both real and the derivation was symmetrical in #u, v#. So we can use one of these values for #u^3# and the other for #v^3# to find the real root:

#x_1 = root(3)((27+3sqrt(69))/54)+root(3)((27-3sqrt(69))/54)#

#color(white)(x_1) = 1/3(root(3)((27+3sqrt(69))/2)+root(3)((27-3sqrt(69))/2))#

#color(white)(x_1) ~~ 1.324717957#

The complex roots are given by multiplying #u# and #v# by suitable powers of #omega=-1/2+sqrt(3)/2i#, the primitive complex cube root of #1#...

#x_2 = 1/3(omega root(3)((27+3sqrt(69))/2)+omega^2 root(3)((27-3sqrt(69))/2))#

#x_3 = 1/3(omega^2 root(3)((27+3sqrt(69))/2)+omega root(3)((27-3sqrt(69))/2))#