How do you expand ${(3 v^{4} + 1)}^{3}$ $(3v^4+1)^3$ ?

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Answer 1 · 2016-07-11T06:24:29

${(3 v^{4} + 1)}^{3} = 27 v^{12} + 27 v^{8} + 9 x^{4} + 1$ $(3v^4+1)^3=27v^12+27v^8+9x^4+1$

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:

enter image source here

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:

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

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

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

How do you expand (3v^4+1)^3(3v4+1)3(3v^4+1)^3?