Question 900270
<pre>
Every term of  {{{(2x-(3/x^2)^"")^6}}}

is of the form

{{{"C(6,k)"(2x)^(6-k)(3/x^2)^k}}}, where k = 0,1,2,3,4,5,6

Write {{{(2x)^(6-k)}}} as {{{2^(6-k)x^(6-k)}}}
Write {{{(3/x^2)}}} as {{{3x^(-2)}}}

{{{"C(6,k)"2^(6-k)x^(6-k)(3x^(-2))^k}}}

Write {{{(3x^(-2))^k}}} as {{{3^k*x^(-2k)}}}

{{{"C(6,k)"2^(6-k)x^(6-k)3^k*x^(-2k)}}}

Rearrange the factors:

{{{"C(6,k)"2^(6-k)3^k*x^(6-k)x^(-2k)}}}

Add the exponents of x

{{{"C(6,k)"2^(6-k)3^k*x^(6-3k)}}}

For the term to be independent of x, the power of x
must be 0, since x<sup>0</sup> = 1 which contains no x.
So we set the exponent of x equal to 0:

{{{6-3k=0}}
{{{-3k=-6}}}
{{{k=2}}}

So we substitute k=2

{{{"C(6,2)"2^(6-2)3^2*x^(6-3*2)}}}

{{{matrix(1,7,"C(6,2)"=15,",",2^(6-2)=2^4=16,",",3^2=9,",",6-3^2=6-6=0)}}}


{{{15*16*9*x^0}}}

{{{2160*1}}}

{{{2160}}}

Edwin</pre>