Question 1117568
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            The correct answer is -28.


            See the full and correct solution below.



<pre>
The binomial expansion is this formula

 
{{{(a+b)^n}}} = {{{a^n}}} + {{{C[n]^1*a^(n-1)*b}}} + {{{C[n]^2*a^(n-2)*b^2}}} + {{{C[n]^3*a^(n-3)*b^3}}} + . . . + {{{C[n]^(n-1)*a^1*b^(n-1)}}} + {{{b^n}}}


In our case,  n = 8,  a = {{{2x^3}}},  b = -{{{1/4x^5}}},  therefore, the binomial expansion in our case is  


{{{(2x^3- 1/(4x^5))^8}}} = {{{(2x3)^8}}} + {{{C[8]^1*(2x^3)^7*(-1/(4x^5))}}} + {{{C[8]^2*(2x^3)^6*(-1/(4x^5))^2}}} + {{{C[8]^3*(2x^3)^5*(-1/(4x^5))^3}}} + . . . + {{{C[8]^9*(2x^3)*(-1/(4x^5))^7}}} + {{{(-1/(4x^5))^8}}}.


The general term of this expansion is  {{{C[8]^k*(2x^3)^(8-k)*(-1/(4x^5))^k}}} = {{{C[8]^k*2^(8-k)*x^(3*(8-k))*(-1/4)^k*(1/x^5)^k}}} = {{{C[8]^k*(-1)^k*2^(8-k-2k)*x^(3*(8-k)-5k)}}}.


The term independent of x is at  3*(8-k) - 5k = 0,  or


    24 - 3k - 5k = 0  ====>  24 = 8k  ====>  k = 3   (the fourth term).


Then the coefficient at this term is  {{{C[8]^3*2^(8-3-2*3)*(-1)}}} = {{{((8*7*6)/(1*2*3))*(-1)*2^(-1)}}} = {{{-(8*7)/2}}} = -28.
</pre>

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The solution by @greenestamps giving the answer  &nbsp;"28"&nbsp; contains an error.