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What is the constant term of the expansion of ?

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What is the constant term of the expansion of $$\left(6x+\dfrac{1}{3x}\right)^6$$?

Creeperhissboom  Apr 16, 2018
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#1
+85726
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Well...let's see!!!

(6x  +  1/ (3x) )^6   =

(6x)^6  +  6 * (6x)^5 (1/ (3x) )  +  15* (6x)^4 (1/(3x))^2  + 20* (6x)^3*(1 /(3x))^3

+ 15(6x)^2 * (1/(3x))^4 +   6* (6x)(1/(3x) )^5  + (1/ (3x))^6

The constant term is in red

20 * 216x^3 / (27x^3)  =  20 *216 / 27   =  20 *24 / 3  =  20 * 8   =  160

CPhill  Apr 17, 2018
#2
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Hey Chris!

Just a quick question, if I do the problem, I will probably hard expand it, but is there are better way that just isolates the constant?

A way that doesn't require you to expand the whole expression?

GYanggg  Apr 17, 2018
#3
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Yeah....GYangg...I just decided to expand it...but..we can see it this way.....

C(6,3)  (6x)^3 (1 / (3x))^3

Notice that the  x^3  terms will cancel   and we will be left with

C(6,3) * 6^3  / 3^3  =

20 * 216 / 27

20 * 8  =

160

CPhill  Apr 17, 2018
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I understand now.

For future references, just call me Gavin, to save you from typing gyanggg :)

GYanggg  Apr 18, 2018
#4
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160 jk i said 160!!!!!!!!

Guest Apr 17, 2018

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