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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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5+0 Answers

 #1
avatar+85726 
+2

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

 

 

cool cool cool

 
CPhill  Apr 17, 2018
 #2
avatar+177 
+2

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
avatar+85726 
+2

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

 

 

cool cool cool

 
CPhill  Apr 17, 2018
 #5
avatar+177 
+1

Thanks for your response. 

 

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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