How do you expand #(3v^2-1)^4#?

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Answer 1 · 2016-09-05T10:20:34

#(3v^2-1)^4=81v^8-108v^6+54v^4-12v^2+1#.

The Binomial Theorem :

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

#+""_nC_3a^(n-3)b^3+............+""_nC_na^(n-n)b^n#.

#:.(a-1)^4=""_4C_0a^(4-0)(-1)^0+""_4C_1a^(4-1)(-1)^1+""_4C_2a^(4-2)(-1)^2+""_4C_3a^(4-3)(-1)^3+""_4C_4a^(4-4)(-1)^4#.

Here, #""_4C_0=""_4C_4=1; ""_4C_1=""_4C_3=4;""_4C_2=6#.

#:. (a-1)^4=a^4-4a^3+6a^2-4a+1#.

Replacing #a" by "3v^2#, we get, #(3v^2-1)^4#

#=(3v^2)^4-4(3v^2)^3+6(3v^2)^2-4(3v^2)^1+1#

#:. (3v^2-1)^4=81v^8-108v^6+54v^4-12v^2+1#.