Question 753595
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The pattern for binomial expansion is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\alpha\ +\ \beta\right)^n\ =\ \sum_{i=0}^n\,{{n}\choose{i}}\,\alpha^{n-i}\beta^i]


For your problem: *[tex \LARGE \alpha\ =\ 2x], *[tex \LARGE \beta\ =\ -3y], and *[tex \LARGE n\ =\ 6].  So the first term, that is when *[tex \LARGE i\ =\ 0], is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ {{6}\choose{0}}\,(2x)^{6-0}(-3y)^0\ =\ (1)(64x^6)(1)\ =\ 64x^6]


Then the second term is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ {{6}\choose{1}}\,(2x)^{6-1}(-3y)^1\ =\ (6)(32x^5)(-3y)\ =\ -576x^5y]


Then the third term is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ {{6}\choose{2}}\,(2x)^{6-2}(-3y)^2\ =\ (15)(16x^4)(-3y)^2\ =\ 2160x^4y^2]



And so on...


You should be able to calculate the remaining 4 terms for yourself. 


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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