# force maximum simplification of radicals in a traditional way-How to do it?

this should be simple but I can't get mathematica to simplify a radical or a set of radicals and express them in the traditional way for example by simplifying this

FullSimplify[(3 a (18 a^4)^(1/4))^3, a > 0]// TraditionalForm (edited)


results in this $$81\ 2^{3/4} \sqrt{3} a^6$$ or this $$27\ 18^{3/4} a^6$$

but not this,

$$81 \sqrt[4]{72} a^6$$

which is the final and most compact

There is some way I can simplify even more and from the results in a traditional way, this is just a small example.

• People here generally like users to post code as Mathematica code instead of just images or TeX, so they can copy-paste it. It makes it convenient for them and more likely you will get someone to help you. You can format inline code and code blocks by selecting the code and clicking the {} button above the edit window. The edit window help button ? is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful Commented Sep 7, 2020 at 4:18
• @Michael E2, Thank you for your indication, the truth is that later I understood your observation, I am new in this forum, and my English is not so good. Commented Sep 7, 2020 at 12:08

FullSimplify[(3 a (18 a^4)^(1/4))^3, a > 0] /.
x_^Rational[y_, z_]*Sqrt[a_] :> (Defer@*Surd)[x^y*a^(z/2), z]


$$81 \sqrt[4]{72} a^6$$

since $$a^{b/c}$$ partly $$=\sqrt[c]{a^b}$$ , so we want to match a,b,c.

But

2^(3/4) /. a_^(b_/c_) :> {a, b, c}


doesn't work.

Because

b/c // AtomQ  (*False*)
3/4 // AtomQ  (*True*)


So mma see 3/4 as a whole.

2^(3/4) // FullForm (*Power[2,Rational[3,4]]*)


Which means we can use

2^(3/4) /. a_^Rational[b_, c_] :> {a, b, c} (*{2,3,4}*)


After that, use Surd and Defer to control the calculation.

• thanks for your contribution, could you clarify a little more how this code works Commented Sep 7, 2020 at 15:58

It is a matter of taste what is simpler. But you can always overwrite MMA decisions by e.g.:

 FullSimplify[(3 a (18 a^4)^(1/4))^3, a > 0] /.
2^(3/4) Sqrt[3] -> HoldForm[72^(1/4)] // TraditionalForm

• ,Thank you, but in this case I arrived at the result with the fourth root of 72 manually, and if I don't know that, if the problem is much more complex, as I demand you to reduce it more and not leave it in terms of high powers to fractional exponents Commented Sep 7, 2020 at 12:06

This combines fractional powers of integers to arrive at the form $$n^{1/d}$$ of a root of an integer. It assumes that the fractional powers came from Power automatically expanding a reduced power.

Simplify[(3 a (18 a^4)^(1/4))^3, a > 0] //.
Power[b1 : (_Integer | _Hold), r1_Rational]*
Power[b2 : (_Integer | _Hold), r2_Rational] :>
With[{d = LCM[Denominator[r1], Denominator[r2]]},
Power[
ReleaseHold[b1]^(d*r1) ReleaseHold[b2]^(d*r2) //
Evaluate // Hold,
1/d]
] /. Hold -> Defer


I wrote DisplayPowersTogether to do this sort of thing.

DisplayPowersTogether=ResourceFunction["DisplayPowersTogether"];
DisplayPowersTogether@FullSimplify[(3 a (18 a^4)^(1/4))^3,a>0]


$$81 \sqrt[4]{72} a^6$$

For ResouceFunctions that you use often, download the source notebook, so you don't have to wait for ResourceFuntion every time. Instead, you can put the code in your init.m file.