SOLUTION: sorry I made a mistake when I typed it in: (4^-2 x y^-3/x^-3 y)^3 times (8^-1 x^-2 y/x^4 y^-1)^-2 = answer x^24/64y^16 - I cannot figure out how to get the 64 in the answer.

Question 164695: sorry I made a mistake when I typed it in:
(4^-2 x y^-3/x^-3 y)^3 times (8^-1 x^-2 y/x^4 y^-1)^-2 = answer x^24/64y^16 - I cannot figure out how to get the 64 in the answer.
Found 2 solutions by Fombitz, Mathtut:
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
Let's take each chunk step by step.
First part:

Second part:

Now let's multiply the first part by the second part:

Let's look at the constant,

There's the 64.

Answer by Mathtut(3670) (Show Source): You can put this solution on YOUR website!
when you cube the 1st term out you get (4^-6*x^3*y^-9/x^-9*y^3) the second term invert and square you get (x^8*y^-2/8^-2*x^-4*y^2) when you switch 8^-2 to the numerator you have 8^2 or 8squared and when you switch 4^-6 to the denominator you get 4^6 or 4*4*4*4*4*4 when you simplify 8^2/4^6 you get 1/64....that is where the 64 comes into play in the answer....
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