Respuesta :

mcherfane05

Answer:

1/(x^3 + 6x^2 + 12x + 8)

Step-by-step explanation:

The first thing we do is rationalize this expression. (2+x)^-3 is written as

1/(2+x)^3

Then from there we can foil out the denominator. It is easiest to foil (2+x)(2+x) first and then multiply that product by (2+x).

(2+x)(2+x) = 4 + 4x + x^2

(4+4x+x^2)(2+x) = 8+8x+2x^2+4x+4x^2+x^3.

Then we combine like terms and put them in order to get:

x^3 + 6x^2 + 12x + 8

And of course we can't forget that this was raised to the negative third power, so our answer is 1/(x^3 + 6x^2 + 12x + 8)

caylus

Answer:

Hello,

Step-by-step explanation:

[tex](a+x)^n=a^n+\left(\begin{array}{c}n\\ 1\end{array}\right)*a^{n-1}*x+\left(\begin{array}{c}n\\ 2\end{array}\right)*a^{n-2}*x^2+\left(\begin{array}{c}n\\ 3\end{array}\right)*a^{n-3}*x^3+\left(\begin{array}{c}n\\ 4\end{array}\right)*a^{n-4}*x^4+...+\left(\begin{array}{c}n\\ n\end{array}\right)*a^{n-n}*x^n[/tex]

[tex]with \\\\\left(\begin{array}{c}n\\ 1\end{array}\right)=n\\\\\left(\begin{array}{c}n\\ 2\end{array}\right)=\dfrac{n(n-1)}{2!} \\\\\left(\begin{array}{c}n\\3 \end{array}\right)=\dfrac{n(n-1)(n-2)}{3!} \\\\...\\[/tex]

[tex]\dfrac{1}{(2+x)^3} =\dfrac{1}{8} +3*\dfrac{x}{4}+3\dfrac{x^2}{2}+x^3\\\\[/tex]

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