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

Can someone please help me i expanded it but I don't know which one is the constant term 


When I expanded it I got this \(46656x^6+15552x^4+2160x^2+160+\frac{20}{3x^2}+\frac{4}{27x^4}+\frac{1}{729x^6} \)

 Nov 7, 2021

Hello. The constant term is the term in which there is no variable. In your case, that is 160. There is no variable multiplying by it. 


Remember: Variables is an unknown value. 


and a constant is a known value with no variables. 

 Nov 7, 2021

Hi JMaster,


I'm giving you one last chance to be polite.   (not just to me)


You should give yourself a point Mathygoo  wink




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


The general term is 


\(\begin{pmatrix} 6\\ r \end{pmatrix} \displaystyle \left(\frac{1}{3x}\right)^r (6x)^{6-r} \)


You need the term where the power of x is 0


so you need to solve this    \(\displaystyle \left(\frac{1}{x}\right)^r (x)^{6-r} =x^0\)


when you solve for r you can then sub that  r  value into the general term to get the constant that you require.

 Nov 7, 2021
edited by Melody  Nov 7, 2021

Thanks Melody

Jmaster10  Nov 7, 2021

Thanks guys it means a lot and that makes so much more sense

 Nov 7, 2021

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