Question

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

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

Answer 1

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`