# For (2m^4-5n^6)^5, what is the third term divided by 2m^2n^{-3}?

## Question:

For {eq}(2m^4-5n^6)^5 {/eq}, what is the third term divided by {eq}2m^2n^{-3} {/eq}?

## Binomial Theorem

The binomial theorem of algebra allows us to expand a polynomial of the form {eq}(x+y)^n {/eq} for any integer n > 0 into a sum of terms. Each term is of the form {eq}px^{q}y^{r} {/eq} with q+r = n and the binomial coefficient {eq}p = \binom{n}{r} {/eq}

We will use the binomial theorem to find the third term of {eq}(2m^4 - 5n^6)^5 {/eq}

According to the binomial theorem: $$(x + y)^n = x^n + \binom{n}{1}x^{n-1}y + \binom{n}{2}x^{n-2}y^2 + ... \binom{n}{n-2}x^2y^{n-2} + \binom{n}{n-1}xy^{n-1} + y^n$$

Plugging in {eq}x = 2m^4, \ y = -5n^6, \ n = 5 {/eq} yields:

$$(2m^4 - 5n^6)^5 = (2m^4)^5 + \binom{5}{1}(2m^4)^4(-5n^6) + \binom{5}{2}(2m^4)^3(-5n^6)^2 + .... + (-5n^6)^5$$

So the third term in the above sum on the right is $$\binom{5}{2}(2m^4)^3(-5n^6)^2 = \frac{5!}{2!3!} * 8m^{12} * 25n^{12} = 2000m^{12}n^{12}$$

Dividing the third term by {eq}2m^2n^{-3} {/eq} yields:

$$\frac{2000m^{12}n^{12}}{2m^2n^{-3}} = 1000m^{10}n^{15}$$

So the answer is {eq}1000m^{10}n^{15} {/eq}.