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# Algebra

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What is the constant term of the expansion of (6x + 1/(3x))^8?

Jul 19, 2021

#1
+63
+2

Alright, let's break this down by simplifying it.

$$(6x+1/3x)^2$$

$$=(6x+\frac{1}{3x})^2$$

Now we can apply the formula

$$(a+b)^2=a^2+2ab+b^2$$

or in this case,

$$(6x + \frac{1}{3x})^2 = 6x^2+2(6x)\frac{1}{3x}+\frac{1}{3x}^2$$

which will become:

6x^2 + 4 + 1/3x2, since 2*6x is 12x, and 12x+1/3x is 12x/3x. Divide by 3x and you get the answer, $$\fbox{4}$$

Jul 19, 2021
#2
+2116
0

Welcome to web2.0calc. :)

Nice solution, but it says raised to the 8th instead of 2nd.

To get the constant term, we need 4 "6x" and 4 "1/(3x)".

Using the binomial theorem, the coefficient is 8C4 = 70.

70 * (6x)^4 * (1/(3x))^4 = 1120

=^._.^=

catmg  Jul 19, 2021
edited by catmg  Jul 20, 2021
#3
+63
+1

Ah. I constantly forget to pay attention to the details. Thanks for the heads up!

PBJcatalinasandwich  Jul 19, 2021