SOLUTION: I have never come across determining which term in the expansion of an equation is a constant, can you help me for the term: (1/(2x^3) - x^5)^8 for which is a constant? thank

Algebra ->  Permutations -> SOLUTION: I have never come across determining which term in the expansion of an equation is a constant, can you help me for the term: (1/(2x^3) - x^5)^8 for which is a constant? thank       Log On


   



Question 647856: I have never come across determining which term in the expansion of an equation is a constant, can you help me for the term:
(1/(2x^3) - x^5)^8 for which is a constant?
thank you much

Found 2 solutions by Edwin McCravy, AnlytcPhil:
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
%281%2F%282x%5E3%29+-+x%5E5%29%5E8

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

%22C%288%2Ck%29%22%2A%0D%0A%0D%0A%281%2F%282x%5E24-k%29%29%2A%28-x%5E5%29%5Ek, where k goes from 0 through 8

 Simplifying











Subtract exponents of x



Simplifying:





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 x0 = 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







56%2A%0D%0A%0D%0A%0D%0A%28%28-1%29%2Ax%5E%280%29%29%2F32+%0D%0A%0D%0A

56%2A%0D%0A%0D%0A%0D%0A%28%28-1%29%2A1%29%2F32+%0D%0A%0D%0A

%2856%2A%28-1%29%29%2F32+

%28-56%29%2F32

-7%2F4

Edwin

Answer by AnlytcPhil(1807) About Me  (Show Source):
You can put this solution on YOUR website!
%281%2F%282x%5E3%29+-+x%5E5%29%5E8

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

%22C%288%2Ck%29%22%2A%0D%0A%0D%0A%281%2F%282x%5E24-k%29%29%2A%28-x%5E5%29%5Ek, where k goes from 0 through 8

 Simplifying











Subtract exponents of x



Simplifying:





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 x0 = 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







56%2A%0D%0A%0D%0A%0D%0A%28%28-1%29%2Ax%5E%280%29%29%2F32+%0D%0A%0D%0A

56%2A%0D%0A%0D%0A%0D%0A%28%28-1%29%2A1%29%2F32+%0D%0A%0D%0A

%2856%2A%28-1%29%29%2F32+

%28-56%29%2F32

-56%2F32

-7%2F4

Edwin