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expand (3z-2b)^5 using Binomial Theorem
(a+b)^n = n on top sigma in middle k=0 to right (n/k)a^(n-k)b^k

(3a-2b)^5 = 5 on top sigma in middle k=0 on bottom(5/k)(3a)^5-k(-2b)^k

(5/0)(3a)^5(-2b)^0+(5/1)(3a)^4(-2b)^1 +(5/2)(3a)^3(-2b)^2

+(5/3)(3a)^2(-2b)^3 + (5/4)(3a)^1(-2b)^4+(5/5)(2a)^0(-2b)^5

=243a^5-810a^4b+1080a^3b^2+-720a^2b^3+240ab^4-32b^5

5!/(5-5)!5!=5!/1-5!=5!/5!1

(5/0)=5!/(5-0)!0!=5!/5!0!=1

please delete last two lines-that was just my work I calculated- it shouldn't e there
actual answer is 243a^5-810a^4b + 1080a^3b^2 -720a^2 b^3 + 240ab^4-32b^5

Don't know if you have to go into sigma notation unless your instructor wants you to do it that way.

I would have in front of me Pascal's Triangle , where I would need the row
1 5 10 10 5 1

so (3a - 2b)^5
= 1(3a)^5 + 5(3a)^4 (-2b) + 10(3a)^3 (-2b)^2 + 10(3a)^2 (-2b)^3 + 5(3a) (-2b)^4 +1(-2b)^5
= 243a^5 - 810a^4 b + 1080a^3 b^2 - 720a^2 b^3 + 240a b^4 - 32b^5

Thank you for checking

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