Have a look at https://socratic.org/help/symbols and particularly note the hash tags at the beginning and end of the example in the text box.
hash (3x-2)(2x+(1/8))^10 hash gives: # (3x-2)(2x+(1/8))^10 #
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#color(blue)("Step 1 - dealing with "(2x+1/8)^10#
To solve this you use the binomial expansion approach.
#color(brown)("This is where Pascal's Triangle comes from.")#
For #(a+b)^10# the #x^7# part is:
# 10" choose "7color(white)("ddd")->color(white)("ddd") a^7b^3xx (10!)/(7!(10-7)!)#
# color(white)("dddddddddddddd")->color(white)("ddd") (2x)^7(1/8)^3xx (10!)/(7!(10-7)!)#
# color(white)("dddddddddddddd")->color(white)("ddd") (128x^7)/512xx (10xx9xx8xx7!)/(7!(3xx2xx1))#
# color(white)("dddddddddddddd")->color(white)("ddd") x^7/cancel(4)^1xx (cancel(10)^5xxcancel(9)^3xxcancel(8)^2xxcancel(7!)^1)/(cancel(7!)^1(cancel(3)^1xxcancel(2)^1xx1))#
Solution for this part #-> 30x^7#
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#color(blue)("Step 2 - dealing with the "3x " from "(3x-2)" part"#
#30x^7xx3x = 90x^8#
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#color(blue)("Step 3 - dealing with the -2 from the "(3x-2)" part"#
I Forgot a bit There is also 2 from (3x-2) times the # ?x^8# part from #(2x+1/8)^10#
Using the above method the #(?x^8)# part from #(2x+1/8)^(10) # is #180x^8#
So combining these we have:
#90x^8+(-2xx180x^8) = -270x^8#
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#color(blue)("Output from Maple")#
You will need to zoom in to see the numbers/structure.
#-270# is confirmed
