how do you do these type of problems??
1. (-125a^5b^-12/27a^2c^6)^-2/3
2. (a^8b^2c^5)/3 square root (a^6b^9c^12)
thank you for your help!
1. Consider each term separately. It also helps if you'd write it clearly so I can tell what's in the numerator and what's in the denominator. I'll take my best guess.
The -2/3 power needs to distribute to every term - the coefficient and all the variables. The variables are easy. You just multiply them by -2/3. The coefficient you just need to solve.
First I'd like to get rid of the fraction.
Becomes -125/27 * a^3 * b^-12 * c^-6 all to the -2/3 power (WHY?)
From here do -125/27 to the -2/3 power. First take the cube root which is -5/3. Then square it to become 25/9. Lastly take the reciprocal because it's a negative exponent. We get 9/25.
Now do the exponents. We get a^-1, b^8, and c^4. So our answer is:
(9 b^8 c^4) / a
==
2. (a^8b^2c^5)/3 square root (a^6b^9c^12)
I'm guessing it needs to be simplified and rationalized. Take the square root first of the denominator (1/2 power) and get a^3*b^(9/2)*c^6. Now just divide to get:
(a^5) / (3*b^(5/2)*c)
Remember that b^(5/2) is the same as b^2*sqrt(b).
