Simplifying expressions on Mathematica is a recurrent topic. However I found myself stuck with this one:

Solving an rational equation on Mathematica gives me the following expression:

expression := -((2 x)/3) + (2^(1/3) (3 - x^2))/(
  3 (9 x - 2 x^3 + Sqrt[4 (3 - x^2)^3 + (9 x - 2 x^3)^2])^(
   1/3)) - (9 x - 2 x^3 + Sqrt[4 (3 - x^2)^3 + (9 x - 2 x^3)^2])^(
  1/3)/(3 2^(1/3))

I know that the expression should simply to $-x$. If I plot the expression I get indeed a plot of $y=-x$.

Plot[expression, {x, -1, 1}]

Resulting plot

I tried using the function "Simplify", "FullSimplify" even "Expand" and then "Simplify" but got no luck as I do not have restriction on $x$. The expression is very messy and I do not manage to simplify it by hand. How could I use mathematica to simplify "expression" to $-x$? (other than using a graphical plot)

Any idea are much appreciated! Thanks

  • 6
    $\begingroup$ Try: Plot[expression, {x, -10, 2}] ? $\endgroup$ Commented Aug 8, 2019 at 17:25
  • 1
    $\begingroup$ How much do you trust magic? In[348]:= First[GroebnerBasis[expression, x]] Out[348]= x $\endgroup$ Commented Jan 6, 2020 at 14:57

1 Answer 1


This isn't the best or most general solution, but you can experiment a bit by Taylor expanding about $x=0$ to some sufficiently high order.

Series[expression, {x, 0, 20}] // Simplify

This results in -x+O[x]^21 which means there are no corrections to $-x$ upto order 21 in the Taylor expansion about $x=0$. If this remains stable to even higher orders you have a probable guess as to what the result is.

  • 2
    $\begingroup$ As @Mariusz Iwaniuk points out the plot isn't a straight line for $x<-2$. This happens clearly because there are branch cuts at $x=\pm2$ from the $\sqrt{4-x^2}$ appearing in the expression. This is obviously not seen if you do a Taylor expansion about $x=0$. $\endgroup$ Commented Aug 8, 2019 at 19:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.