How do I find the constant term from a given binomial without expanding the whole expression?

#(x - 1/x^(1/3))^12#

1 Answer
MathFact-orials.blogspot.com
Jul 17, 2018

See below for an idea:

Explanation:

The constant term in a binomial expansion is the one, if it exists, that has no variable terms within it.

This prior answer on Socratic has a great explanation on how to find the constant term:

How do I use the binomial theorem to find the constant term?

Following the instructions in that answer, we first set up the general term:

#T_(r+1)=((12),(r))x^(12-r)(-1/x^(1/3))^r#

and now simplifying:

#T_(r+1)=((12),(r))x^(12-r)(-1)^r(x^(-r/3))#

#T_(r+1)=((12),(r))x^(12-r-(r/3))(-1)^r#

#T_(r+1)=((12),(r))x^(12-(4r)/3)(-1)^r#

The constant term is the one that will have #x=0#, so we have:

#x^(12-(4r)/3)=x^0#

giving

#12-(4r)/3=0#

#12=(4r)/3#

#4r=36=>r=9#

Therefore, the 9th term in the expansion is the one with no #x# terms within it:

#((12),(9))x^3(-1/x^(1/3))^9#

and to show that (i.e. checking our work):

#220xxx^3xx(-1/x^3)=-220#

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