{(a^ -1 b^2/a^2 b^ -4)^7} / {a^3 b^ -5 /a^ -2 b^3)^ -5}?

Question

its a question of maths(surds n indices).....

Answer 1

let's take the numirator {(a^ -1 b^2/a^2 b^ -4)^7} :
first:
(a^ -1 b^2/a^2 b^ -4) = a^(-1-2) * b^(2--4) = a^-3 * b^6
all of this ^7 = a^-21 * b^42

now for the denominator {a^3 b^ -5 /a^ -2 b^3)^ -5}
first:
(a^3 b^ -5 /a^ -2 b^3) = a^(3--2) * b^(-5-3) = a^5 * b^-8
all of this^ -5 = a^-25 * b^40

(a^-21 * b^42) /( a^-25 * b^40) = a^(-21--25) * b^(42-40)
= a^3 *b^2

Check it out with any values of a and b and it will work, guaranteed.

Answer 2

(((a^-1 * b^2)/(a^2 * b^-4))^7) / (((a^3 * b^-5)/(a^-2 * b^3))^-5)
(a^(-1 - 2) * b^(2 - (-4))^7 / (a^(3 - (-2)) * b^(-5 - 3))^-5
(a^(-1 + (-2)) * b^(2 + 4))^7 / (a^(3 + 2) * b^(-5 + (-3))^-5
(a^-3 * b^6)^7 / (a^5 * b^-8)^-5
(a^-3 * b^6)^7 * (a^5 * b^-8)^5
(a^(-3 * 7) * b^(6 * 7)) * (a^(5 * 5) * b^(-8 * 5))
(a^-21 * b^42)(a^25 * b^-40)
a^(-21 + 25) * b^(42 + (-40))
a^(-21 - (-25)) * b^(42 - 40)
a^4b^2

ANS : a^4b^2

Answer 3

well can u answer this question since ur so smart is that a homework question that u cant do and want some else to do it 4 u

Answer 4

THIS IS NOT A DIFFICULT OPERATION, REMEMBER, WHEN YOU MULTIPLY YOU ADD THE INDICES, WHEN YOU DIVIDE, YOU SUBSTRACT, WHEN YOU GIVE A POWER, YOU MILTIPLY COEFFICIENTS AND WHEN YOU GET THE ROOT, YOU DIVIDE.
AND START FROM INSIDE TO OUTSIDE
THE ANSWER MUST BE a^4b^2

Answer 5

did all this on a seperate sheet of paper. I promise! Don't have time to write out all steps.

The answer is a^-8 b^20

Hope this helps

Answer 6

{(a^-1b^2/a^2b^-4)^7}/{a^3b^-5/a^-2b^3)^-5}

(a^-7b^14/a^14b^-28)/(a^-15b^25/a^10b^-15)

(a^-7b^14/a^14b^-28) X (a^10b^-15/a^-15b^25)

a^(-7-14+10+15)b^(14-15+28-25)

a^4b^2

Answer 7

did u make this up, or is it a home work problem?

Answer 8

uhm if u figure that out..let me noe