Simplify.(Is this problem more clear than the previous post?)
First part..
a^4/3 times c^1/6 all over 9b^3...everything is in the parenthesis and is raised to the power of -2
MULTIPLY TO THIS SECOND SET:
a^3 times b^2 all over 3c.. everything is in the parenthesis and is raised to the power of -1/4
If you have
[(a^4/3 * c^1/6) / (9b^3)]^-2
That's (81b^6) / (a^8/3 * c^1/3)
[(a^3 * b^2)/(3c)]^-1/4 is
(3^1/4 * c*1/4) / (a^3/4 * b^1/2)
multiply to get
(81*3^1/4 * b^11/2) / (a^41/12 * c^1/12)
messy
To simplify the given expressions, we can use the rules of exponents. Let's break it down step by step:
First expression:
(a^(4/3) * c^(1/6))^(-2) / (9b^3)^(-2)
- When we raise a power to another power, we multiply the exponents:
a^((4/3)*(-2)) * c^((1/6)*(-2)) / (9b^3)^(-2)
- Simplifying the exponents:
a^(-8/3) * c^(-1/3) / (9^(-2) * b^(-6))
- Simplifying the base with negative exponents:
1 / (a^(8/3) * c^(1/3) * 9^2 * b^6)
- Rearranging the terms:
1 / (9^2 * a^(8/3) * b^6 * c^(1/3))
Second expression:
(a^3 * b^2 / 3c)^(-1/4)
- Applying the negative exponent:
1 / (a^3 * b^2 / 3c)^(1/4)
- Taking the fourth root of the fraction:
1 / (a^(3/4) * b^(1/2) * (3c)^(1/4))
Now we have simplified both expressions separately. If you have any specific operation you would like to perform with these expressions, please let me know.