How do you expand (3v^2-1)^4(3v2−1)4?

1 Answer
Ratnaker Mehta
Sep 5, 2016

(3v^2-1)^4=81v^8-108v^6+54v^4-12v^2+1(3v2−1)4=81v8−108v6+54v4−12v2+1.

Explanation:

The Binomial Theorem :

(a+b)^n=""_nC_0a^(n-0)b^0+""_nC_1a^(n-1)b^1+""_nC_2a^(n-2)b^2(a+b)n=nC0an−0b0+nC1an−1b1+nC2an−2b2

+""_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.

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