# Replacement rules do not apply for an expression inside a root

Say, you have an expression which looks like this:

myexpr = (mess)+(some other mess)/( (mess)^3+(some other mess)^2 )^(1/3)

Here the 'mess' in the name of terms stands for what it is - a mess. To make it look nicer I use replacement rules and substitute (mess)->u, (some other mess)->v as follows

myexpr /. {(mess)->u,(some other mess)->v}

so that the output would look as nice as

u+v/(u^3+v^2)^(1/3)

Now, I actually don't get this result because when done as above the rules are not applied inside the cubic root, so the output I get instead is

u+v/( (mess)^3+(some other mess)^2 )^(1/3)

The question is, naturally, how can the full replacement be achieved with minimum pain? Thanks.

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• We will probably need to see the messes you are dealing with to help you find your problem. Sometimes problems like this are caused by a rule being applied to an outer expression, and not to an inner one. You might look at ReplaceRepeated – mfvonh Aug 8 '15 at 12:52

Your have to use RuleDelayed (:>) in such a case, so that the evaluation is not done immediately:

myexpr = (mess) + (someOtherMess)/((mess)^3 + (someOtherMess)^2)^(1/3)

and then:

myexpr /. {mess :> x, someOtherMess :> y}

Then you get what you want.

• @Patrick: Thanks for removing my typo - I didn´t see it.. – mgamer Aug 8 '15 at 15:14
• I'm not sure that I follow your reasoning here. RuleDelayed will delay the evaluation of $x$ and $y$ in your example, but won't affect the (mess) expressions. Since I expect that $x$ and $y$ might not even have been assigned a value in this case, I don't see the difference between this and a simple Rule. The opposite behavior might be required, i.e. we might have to prevent evaluation of mess with one of the Hold* functions so it can be matched literally to what is in the original expression. Of course, it's hard to be specific without seeing the OP's actual expressions. – MarcoB Aug 8 '15 at 15:24
• I agree. In most cases ReplaceRepeated would be the right choice. But without having the original code, it´s hard to find an appropriate solution – mgamer Aug 8 '15 at 16:03

Thanks to mfvonh for the answer - using ReplaceRepeated (//.) as shown below replaces everything.

myexpr //. {(mess)->u,(some other mess)->v}

To those in power: I unfortunately cannot accept the answer properly because after setting up the account the question is not listed as mine (and can't leave a comment because low rep = having no rights).