Question

Find the constant term in the expansion of (2x^2-[1/x])^6

Answer 1

\(\large\color{#000000 }{ \displaystyle\left(ax^2+b/x\right)^n }\)

\(\large\color{#000000 }{ \displaystyle\left({~[ax^3+b]\cdot [1/x]~}\right)^n }\)

\(\large\color{#000000 }{ \displaystyle \left(1/x\right)^n\left(ax^3+b\right)^n }\)


So,

\(\large\color{#000000 }{ \displaystyle\left(ax^3+b\right)^n }\)
Will have one constant term,

since everything is multiplied times
\(\large\color{#000000 }{ \displaystyle \left(1/x\right)^n=x^{-n} }\)

You need to find the term of
\(\large\color{#000000 }{ \displaystyle\left(ax^3+b\right)^n }\)
that will have a power of n
(so that the product of powers gives a 0 power)

Answer 2

In your case,

\(\large\color{#000000 }{ \displaystyle\left(2x^2+1/x\right)^6 }\)

\(\large\color{#000000 }{ \displaystyle\left({~[2x^3+1]\cdot [1/x]~}\right)^6 }\)

\(\large\color{#000000 }{ \displaystyle \left(1/x\right)^6\left(2x^3+1\right)^6 }\)


So,

\(\large\color{#000000 }{ \displaystyle\left(2x^3+1\right)^6 }\)
Will have one constant term,

since everything is multiplied times
\(\large\color{#000000 }{ \displaystyle \left(1/x\right)^6=x^{-6} }\)

You need to find the term of
\(\large\color{#000000 }{ \displaystyle\left(2x^3+1\right)^6 }\)
that will have a power of 6
(so that the product of powers gives a 0 power)

Answer 3

My route seems ong, but you just need to find a single term from (2x^3+1)^6 that will have a power (of x) of 6, and then everything is multiplied times x^(-6), so multiply this term times x^(-6) and you get your constant term.

Answer 4

You can use pascal's triange.

Answer 5

In terms of a and b, I find that this term with power of 6, is:

\(\large\color{#000000 }{ \displaystyle 15a^2b^4x^6 }\)