Question 647856
<pre>
{{{(1/(2x^3) - x^5)^8}}}

There are 9 terms in the expansion, each one is of this form:

{{{"C(8,k)"*

(1/(2x^24-k))*(-x^5)^k}}}, where k goes from 0 through 8

 Simplifying

{{{"C(8,k)"*


(1/(2^(8-k)*x^(24-k))))*(-x)^(5k)}}}

{{{"C(8,k)"*


(1/(2^(8-k)*x^(24-3k)))*((-1)*x)^(5k)}}}

{{{"C(8,k)"*


(1/(2^(8-k)*x^(24-3k)))*((-1)^(5k)*x^(5k))}}}

{{{"C(8,k)"*

(1/(2^(8-k)*x^(24-3k)))*(((-1)^(5k)*x^(5k))/1)}}}

{{{"C(8,k)"*
(((-1)^(5k)*x^(5k))/(2^(8-k)*x^(24-3k)))}}}

Subtract exponents of x

{{{"C(8,k)"*


((-1)^(5k)*x^(5k-(24-3k)))/( 2^(8-k) )

}}}

Simplifying:

{{{"C(8,k)"*


((-1)^(5k)*x^(5k-24+3k))/( 2^(8-k) )

}}}

{{{"C(8,k)"*


((-1)^(5k)*x^(8k-24))/( 2^(8-k) )

}}}

The only time x raised to a power is NOT a variable, but a
constant, is when the exponent of the power is 0, since x<sup>0</sup> = 1,
and 1 is a constant, not a variable.


Therefore we take the exponent of x, which is 8k-24, and set it
 equal to 0:

8k-24 = 0
   8k = 24
    k = 3

So we substitute k = 3 in

{{{"C(8,k)"*


((-1)^(5k)*x^(8k-24))/( 2^(8-k) )

}}}

{{{"C(8,3)"*


((-1)^(5*3)*x^(8*3-24))/( 2^(8-3) )

}}}

{{{56*


((-1)^(15)*x^(24-24))/( 2^(5) )

}}}

{{{56*


((-1)*x^(0))/32 

}}}

{{{56*


((-1)*1)/32 

}}}

{{{(56*(-1))/32 }}}

{{{(-56)/32}}}

{{{-7/4}}}

Edwin</pre>