Recall a binomial series follows:

`(1+x)^k=sum_(n=0)^oo _(k(k-1)(k-2)...(k-n+1))/(n!)x^n`

or

`(1+x)^k = 1 + kx + (k(k-1))/(2!) x^2 + (k(k-1)(k-2))/(3!)x^3 +(k(k-1)(k-2)(k-3))/(4!)x^4+` ...

To evaluate given function `f(x) =1/(2+x)^3` , we may apply `2+x = 2(1+x/2)` .

The function becomes:

`f(x) =1/(2(1+x/2))^3`

Apply Law of Exponents: `(x*y)^n = x^n*y^n` at the denominator side.

...

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Recall a binomial series follows:

`(1+x)^k=sum_(n=0)^oo _(k(k-1)(k-2)...(k-n+1))/(n!)x^n`

or

`(1+x)^k = 1 + kx + (k(k-1))/(2!) x^2 + (k(k-1)(k-2))/(3!)x^3 +(k(k-1)(k-2)(k-3))/(4!)x^4+` ...

To evaluate given function `f(x) =1/(2+x)^3` , we may apply `2+x = 2(1+x/2)` .

The function becomes:

`f(x) =1/(2(1+x/2))^3`

Apply Law of Exponents: `(x*y)^n = x^n*y^n` at the denominator side.

`1/(2(1+x/2))^3=1/(2^3(1+x/2)^3)`

`= 1/(8(1+x/2)^3)`

Apply Law of Exponents: `1/x^n = x^(-n)` .

`f(x) = 1/8(1+x/2)^(-3)`

Apply the binomial series on `(1+x/2)^(-3)` . By comparing "`(1+x)^k` " with "`(1+x/2)^(-3)` " the corresponding values are:

`x=x/2 ` and` k =-3`

Then,

`(1+x/2)^(-3) =sum_(n=0)^oo _((-3)(-3-1)(-3-2)...(-3-n+1))/(n!)(x/2)^n`

` =1 + (-3)x/2 + ((-3)(-3-1))/(2!) (x/2)^2 + ((-3)(-3-1)(-3-2))/(3!)(x/2)^3 +((-3)(-3-1)(-3-2)(-3-3))/(4!)(x/2)^4+...`

` =1 -(3x)/2 + ((-3)(-4))/(2!) (x^2/4) + ((-3)(-4)(-5))/(3!)(x^3/8) +((-3)(-4)(-5)(-6))/(4!)(x^4/16)-` ...

` =1 -(3x)/2 +12/(2!) (x^2/4) -60/(3!)(x^3/8) +360/(4!)(x^4/16)-` ...

`=1 -(3x)/2 +(3x^2)/2 -(5x^3)/4 +(15x^4)/16-` ...

Applying `(1+x/2)^(-3) =1 -(3x)/2 +(3x^2)/2 -(5x^3)/4 +(15x^4)/16-` ..., we get:

`1/8(1+x/2)^(-3)=1/8*[1 -(3x)/2 +(3x^2)/2 -(5x^3)/4 +(15x^4)/16-...]`

`=1/8-(3x)/16 +(3x^2)/16 -(5x^3)/32 +(15x^4)/128-` ...

Therefore, the Maclaurin series for the function `f(x) =1/(2+x)^3` can be expressed as:

`1/(2+x)^3=1/8-(3x)/16 +(3x^2)/16 -(5x^3)/32 +(15x^4)/128-` ...