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It seems very simple question, I wish to get rid of the higher order by product of a*b. Any help is appreciated!

f = a^4 + 4 a^3 b + 6 a^2 b^2 + 4 a b^3 + b^4
f /. a b -> 0
(*output*)
a^4 + 4 a^3 b + 6 a^2 b^2 + 4 a b^3 + b^4
(*I want to have*)
a^4+b^4
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3 Answers 3

Reset to default
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Looking at FullForm, it seems this pattern will do it

pat0 = Times[_. , a | Power[a, _], b | Power[b, _]]

Examples:

f = a^4 + 4 a^3 b + 6 a^2 b^2 + 4 a b^3 + b^4;
f1 = 3 (a b - 2 c);
f2 = (a b - 2 c)/(c - a b)
DeleteCases[f, pat0, Infinity]
DeleteCases[f1, pat0, Infinity]
DeleteCases[f2, pat0, Infinity]

Mathematica graphics

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  • $\begingroup$ Thank you for the answer! Can I ask one more case? it works nicely with no denominator, but with it ,(a b - 2 c )/(c - a b), it does not work.How can I do it for the denominator? $\endgroup$
    – Saesun Kim
    Sep 3, 2016 at 22:31
  • $\begingroup$ @SaesunKim added option Infinity. This seems to work now on the new expression you added. Infinity tells it to keep searching for the pattern. $\endgroup$
    – Nasser
    Sep 3, 2016 at 23:17
  • $\begingroup$ Thank you so much. It is really amazing! I really appreciate your help :D $\endgroup$
    – Saesun Kim
    Sep 3, 2016 at 23:26
  • $\begingroup$ Wouldn't it be nice if your solution works on the equivalent expression f=(a + b)^4 ? $\endgroup$
    – yarchik
    Sep 4, 2016 at 17:36
  • $\begingroup$ @yarchik it will work, but need to expand first. This works: DeleteCases[Expand[(a + b)^4], pat0, Infinity] gives a^4+b^4. $\endgroup$
    – Nasser
    Sep 5, 2016 at 8:20
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I would use an algebraic function for an algebraic operation, similar to what I did in the similar question, Replacement rules and algebra:

PolynomialMod[f, a b]
(*  a^4 + b^4  *)

Works with the rational function from a comment by the OP:

Block[{f = (a b - 2 c)/(c - a b)},
 PolynomialMod[f, a b]
 ]
(*  -2  *)
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  • $\begingroup$ Thank you for introducing cool command! I can see that polynomialMod can be really useful! $\endgroup$
    – Saesun Kim
    Sep 5, 2016 at 17:36
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$\begingroup$ 
 f /. a^(_:1) b^(_:1) :> 0

a^4 + b^4

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1
  • $\begingroup$ (a + b)^4 /. a^(_ : 1) b^(_ : 1) :> 0 yields the same expression... $\endgroup$
    – yarchik
    Sep 4, 2016 at 17:37

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