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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

#1
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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
#2
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Hi JMaster,

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

You should give yourself a point Mathygoo

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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
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Thanks Melody

Jmaster10  Nov 7, 2021
#3
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Thanks guys it means a lot and that makes so much more sense

Nov 7, 2021