(12x^4y^-2/4x^5y^2)^-3? explain?

Question

The answer is 1/27x^3y^12 but i don't get it.
Shouldn't the Y cancel out? so it would be (3/x)^-3?
Can someone explain please?

Answer 1

[(12 x^4 y^-2)/(4 x^5 y^2)]^-3

Simplify within the brackets by combining like terms.
Remember that when dividing two terms, you want to subtract the exponents:
[3 x^(4-5) y^(-2-2)]^(-3)

Simplify the exponents to get:
[3 x^(-1) y^(-4)]^(-3)

Distribute the exponent into each factor:
3^(-3) x^(-1*-3) y^(-4*-3)

Simplify the exponents once more to get:
(1/27) x^3 y^12

Answer 2

[{12(x^4)(y^-2)} / {4(x^5)(y^2)}]^-3

OK, the reason the y doesn't cancel out is because y^-2 != y^2. The first thing I would do is reverse the signs on the denominator exponents. Doing this basically puts everything under one (by definition: z^-1 = 1/z) so you don't have to deal with a fraction. That means {(4^1)(x^5)(y^2)} becomes {(1/4)(x^-5)(y^-2)} (remember that 4^-1 = 1/4). Rearranging so like terms are close gives you
{(12)(1/4)*(x^4)(x^-5)*(y^-2)(y^-2)} ^ -3
Because (z^a)(z^b) = z^(a+b), we can combine like terms, giving us...
{ (3) * x^(4 + -5) * y^(-2 + -2) } ^-3
{ 3 (x^-1) (y^-4)} ^-3
Now, because (abc)^z = (a^z)(b^z)(c^z), and (a^b)^c = a^(bc), we can go forward...
(3^-3) [x^(-1*-3)] [y^(-4*-3)]
[1/27][x^3][y^12]

Another way of doing this problem is to get rid of all the negative exponents inside the parenthesis, and simplify from there...

{12(x^4) / 4(x^5)(y^2)(y^2)} ^ -3
{12 (x^4) / 4(x^5)(y^4)} ^ -3
{3 / (x)(y^4) } ^-3
{27 / (x^-3)(y^-12)}
{(x^3) * (y^12) / 27 }

Of course, you should probably note somewhere that x != 0, y != 0; because you can't divide by 0.

Answer 3

((12x^4y^-2)/(4x^5y^2))^-3

((12/4) * x^(4 - 5) * y^(-2 - 2))^-3

(3x^(-1) * y^(-2 + (-2)))^-3

(3x^(-1) * y^(-4))^-3

(3^-3) * x^(-1 * (-3)) * y^(-4 * (-3))

(1/27)x^3y^12