SOLUTION: Which term in the expansion of (1/(2x^3) - x^5))^8 is a constant? I know you expand it into the equation = (nCk) a^(n-k) b^k to get = 8Ck (1/(2x^3))^(8-k) (-x^5)^k bu

Algebra ->  Permutations -> SOLUTION: Which term in the expansion of (1/(2x^3) - x^5))^8 is a constant? I know you expand it into the equation = (nCk) a^(n-k) b^k to get = 8Ck (1/(2x^3))^(8-k) (-x^5)^k bu      Log On


   

Question 927431: Which term in the expansion of (1/(2x^3) - x^5))^8 is a constant?
I know you expand it into the equation

= (nCk) a^(n-k) b^k
to get = 8Ck (1/(2x^3))^(8-k) (-x^5)^k
but I am not sure what to do after that.
Thank you for the help!
Found 2 solutions by richard1234, Edwin McCravy:
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
The constant term (i.e. x^0 coefficient) is zero. To see it, note that the expression is equal to so all of the terms with nonzero coefficients are divisible by x^3.

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
I don't think the "bee" above has "buzzed" it as clearly
as he could have.J

Although he did say essentially this, which is the whole crux of the
matter:

For a variable to become a constant, 
it must be raised to the 0 power.

8Ck%2A%281%2F%282x%5E3%29%29%5E%288-k%29%2A%28-x%5E5%29%5Ek

Write 1%2F%282x%5E3%29 with a negative exponent, %282x%5E3%29%5E%28-1%29

8Ck%2A%28%282x%5E3%29%5E%28-1%29%29%5E%288-k%29%2A%28-x%5E5%29%5Ek

Multiply exponents and write %282x%5E3%29%5E%28-1%29 as 2%5E%28-1%29x%5E%28-3%29 

8Ck%2A%282%5E%28-1%29x%5E%28-3%29%29%5E%288-k%29%2A%28-x%5E5%29%5Ek

Multiply exponents again and write %282%5E%28-1%29x%5E%28-3%29%29%5E%288-k%29 as 2%5E%28-8%2Bk%29x%5E%28-24%2B3k%29

8Ck%2A2%5E%28-8%2Bk%29x%5E%28-24%2B3k%29%2A%28-x%5E5%29%5Ek

Write %28-x%5E5%29%5Ek as %28-1%2Ax%5E5%29%5Ek and then as %28-1%29%5Ek%2Ax%5E%285k%29

8Ck%2A2%5E%28-8%2Bk%29x%5E%28-24%2B3k%29%2A%28-1%29%5Ek%2Ax%5E%285k%29

Next add the exponents of x: -24%2B3k%2B5k and get -24%2B8k

8Ck%2A2%5E%28-8%2Bk%29%28-1%29%5Ek%2Ax%5E%28-24%2B8k%29

Now remember what I said in the beginning.
To becom a constant the variable x must be raised to the 0 power:

So we set the exponent of x equal to 0

-24%2B8k=0
8k=24
k=3

So we substitute 3 for x:

8Ck%2A2%5E%28-8%2Bk%29%28-1%29%5Ek%2Ax%5E%28-24%2B8k%29
8C3%2A2%5E%28-8%2B3%29%28-1%29%5E3%2Ax%5E%28-24%2B8%2A3%29
56%2A2%5E%28-5%29%28-1%29%2Ax%5E%28-24%2B24%29
56%281%2F2%5E5%29%28-1%29%2Ax%5E0
56%281%2F32%29%28-1%29%2A1
-56%2F32
-7%2F4

For your information so as to see what the entire expansion
looks like when simplified:

1%2F%28256x%5E24%29+%22%22-%22%22%0D%0A%0D%0A1%2F%2816x%5E16%29+%22%22%2B%22%22%0D%0A%0D%0A7%2F%2816x%5E8%29+%22%22-%22%22%0D%0A%0D%0A7%2F4++%22%22%2B%22%22%0D%0A%0D%0A35x%5E8%2F8++%22%22-%22%22%0D%0A%0D%0A7x%5E16+%22%22%2B%22%22%0D%0A%0D%0A7x%5E24++%22%22-%22%22%0D%0A%0D%0A4x%5E32+%22%22%2B%22%22%0D%0A%0D%0A%0D%0Ax%5E40 

So the constant term is really the 4th term, when k=3.

Edwin