how do you simplify this fraction???

Question

( (4ac)/b^-3 ) ^-2

the answer is 1/16a^2b^6c^2

but i can't get it, can someone explain?


also....

(2√3-√5)^2

the answer is 17-4√15


please explain! thx

Answer 1

Question 1
(4ac b³)^(-2)
= 4^(-2) (a^(-2)) (c^(-2)) / b^6
= 1 / (16 a² c² b^6)

Question 2
(2√3 - √5) (2√3 - √5)
12 - 2 √3√5 - 2√3√5 + 5
12 - 4√3√5 + 5
17 - 4√15

Answer 2

#1. The "trick" to this is the reciprocal rule.
It states that if you have (A/B)^-C, this is equal to
(B/A)^C. Applying that to this problem , we get
[b^-3 / 4ac ]^2. However, we can invoke the rule again: b^-3/1 = 1/b^3. Then we wind up with
[1/4acb^3]^2, which when expanded, is the answer.

In the second problem (a-b)^2 = a^2+b^2-2ab
So using 2sqrt(3) as a and sqrt(5) as b, we have
12 + 5 - 2 (2sqrt(3)*sqrt5), which simplifies to the answer.

Answer 3

1)
( (4ac)/b^-3 ) ^-2
= (b^-3/4ac)^2
= (1/4acb^3)^2
= 1/16a^2 c^2 b^6

2)
(2√3-√5)^2
= (2√3)^2 - 4√15 + (√5)^2
= 4*3 - 4√15 + 5
= 12 - 4√15 + 5
= 17 - 4√15

Answer 4

Get rid of the pesky negative powers first.
*since the b has a negative 3 power, you'll want to move it upstairs with the 4ac. when you do that, it's power becomes positive.
*the negative 2 power on the outside tells you to 'flip' the fraction. (that is, find it's inverse ... 4acb^3 becomes "one OVER 4acb^3), and the 2nd power on the outside becomes positive

(1 / 4acb^3)^2

the answer falls out imediately

Your second problem is way different. Rewrite the problem as
(2*sqrt 3 - sqrt 5)(2*sqrt 3 - sqrt 5)
and FOIL.

Wow, I need to learn how to make the little square root signs.
:-)