How do you expand the binomial #(x^3+3)^5# using the binomial theorem?

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Answer 1 · 2016-07-21T18:16:47

#x^15 + 15x^12 + 90x^9 + 270x^6 + 405x^3 + 243#

Binomial theorem:

#(a+b)^n = sum_(k=0)^n ((n),(k)) a^(n-k)b^k#

#(x^3+3)^5 = sum_(k=0)^5 ((5),(k)) (x^3)^(5-k)3^k#

#=((5),(0))(x^3)^5 3^0 + ((5),(1))(x^3)^4 3^1 + ((5),(2))(x^3)^3 3^2 + ((5),(3))(x^3)^2 3^3 + ((5),(4))(x^3)^1 3^4 + ((5),(5))(x^3)^0 3^5#

Remember that #((u),(v)) = (u!)/(v!(u-v)!)#

#therefore 1*x^15 + 5*x^12*3 + 10*x^9*9 + 10*x^6*27 + 5*x^3*81+1*243#

Expansion is:

#x^15 + 15x^12 + 90x^9 + 270x^6 + 405x^3 + 243#