How do you expand #(3v^4+1)^3#?

1 Answer
George C.
Jul 11, 2016

#(3v^4+1)^3=27v^12+27v^8+9x^4+1#

Explanation:

From the binomial theorem, we have the general formula:

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

where #((n),(k)) = (n!)/(k!(n-k)!)#

These coefficients occur as rows in Pascal's triangle:

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For #n=3#, we pick the row #1, 3, 3, 1# to find:

#(a+b)^3 = ((3),(0))a^3+((3),(1))a^2b+((3),(2))ab^2+((3),(3))b^3#

#color(white)(XXXX)=a^3+3a^2b+3ab^2+b^3#

#color(white)()#
For our example, #a=3v^4# and #b=1#, so we find:

#(3v^4+1)^3=(3v^4)^3+3(3v^4)^2+3(3v^4)+1#

#color(white)(XX)=3^3v^12+3*3^2v^8+3*3x^4+1#

#color(white)(XX)=27v^12+27v^8+9x^4+1#

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