Which term in the expansion of ((1/2x^3)-(x^5))^8 is a constant?

THANKS!!

1

x is variable

3,5,8 is exponents

noo...it's asking after u expand the entire thing, which term is a constant ....for example

(2a-3)^2=
(2a-3)(2a-3)
4a^2-12a+9

Then the third term would be "constant" cause it doesn't have any variables attached to it...

To find the constant term in the expansion of ((1/2x^3)-(x^5))^8, we can use the Binomial Theorem. The Binomial Theorem states that for any two terms, (a + b)^n, the constant term can be found using the combination formula:

C(n, r) * a^(n-r) * b^r

Where C(n, r) represents the binomial coefficient, also known as "n choose r".

In our case, the expression ((1/2x^3)-(x^5))^8, the term (1/2x^3) is represented by 'a' and the term (-x^5) is represented by 'b'.

We want to find the constant term when we expand ((1/2x^3)-(x^5))^8, so we want to find the term where r = 8 - r, or equivalently, r = 4.

Plugging the values into the formula, we have:

C(8, 4) * (1/2x^3)^(8-4) * (-x^5)^4

C(8, 4) = 8! / (4! * (8-4)!) = 70

(1/2x^3)^(8-4) = (1/2x^3)^4 = (1/16x^12)

(-x^5)^4 = x^20

Putting everything together:

70 * (1/16x^12) * x^20 = 70/16x^(-12) * x^20 = 70/16x^(-12 + 20) = 70/16x^8 = 35/8x^8

Therefore, the term that represents the constant in the expansion of ((1/2x^3)-(x^5))^8 is 35/8x^8.