How do you expand the binomial #(x^3+y^2)^3# using the binomial theorem?

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Answer 1 · 2016-09-22T13:24:01

#(x^3+y^2)^3=x^9+3x^6y^2+3x^3y^4+y^6#.

#=""_nC_0a^(n-0)*b^0+""_nC_1a^(n-1)b^1+""_nC_2a^(n-2)b^2+...+""_nC_na^(n-n)b^n#.

#:. (a+b)^3=""_3C_0a^(3-0)b^0+""_3C_1a^(3-1)b^1+""_3C_2a^(3-2)b^2+""_3C_3a^(3-3)b^3#.

Here, #""_3C_0=""_3C_3=1, ""_3C_1=""_3C_2=3#. Hence,

#(a+b)^3=a^3+3a^2b+3ab^2+b^3#.

Now, letting, #a=x^3, and, b=y^2#, we have,

#(x^3+y^2)^3=(x^3)^3+3(x^3)^2(y^2)+3(x^3)(y^2)^2+(y^2)^3, i.e.,#

#(x^3+y^2)^3=x^9+3x^6y^2+3x^3y^4+y^6#.