Need math help! (math magicians only!)

i need to simplify using the law of exponents then evaluate:

a) (3 x^4)^9 / (3^5 x^6)^3

b) (6^3 a^5 / 5^5 b^9)^7 x (5^11 b^2 / 6^6 a^3)^3


i need to make sure my answers are right before i hand them in so please answer these questions so i can correct mine

law of exponents; to the power of a power, multiply the exponents.

[3^(1times9) TIMES x^(4times9)] / [3^(5times3) TIMES x^(6times3)]

[3^9 x^36] / [3^15 x^18]

law of exponents; divdie bases, subtract exponents

3^(9-15) times x^(36-18)

3^-6 times x^18

law of exponents; to the power of a negative, flip the fraction and make power positive.

(1/3)^6 times (x^18)/1

1^6(x^18) / 3^6(1)

x^18 / 3^6

x^18 / 729





[6^3a^5 / 5^5b^9]^7 TIMES [5^11b^2 / 6^6a^3]^3

to the power of a power, multiply the exponents

[6^(3times7)a^(5times7) / 5^(5times7)b^(9times7)] TIMES [5^(11times3)b^(2times3) / 6^(6times3)a^(3times3)]

[6^21a^35 / 5^35b^63] TIMES [5^33b^6 / 6^18a^9]

do some cross cancelling.

NOTE that 6^18 goes into 6^21 with 6^3 left over, and into 6^18 with nothing left over

[6^3a^35 / 5^35b^63] TIMES [5^33b^6 / a^9]

NOTE that a^9 goes into a^35 with a^26 left over, and into a^9 with nothing left over

[6^3a^26 / 5^35b^63] TIMES [5^33b^6 / 1]

NOTE that 5^33 goes into 5^35 with 5^2 left over, and into 5^33 with nothing left over

[6^3a^26 / 5^2b^63] TIMES [b^6 / 1]

NTOE that b^6 goes into b^63 with b^57 left over and into b^6 with nothing left over

(6^3a^26 / 5^2b^57) TIMES 1/1

6^3a^26 / 5^2b^57

216a^26 / 25b^57

can't simplify further

(3 x^4)^9 / (3^5 x^6)^3=(3^9x^36)/(3^15x^18)
=x^18/3^6
(6^3 a^5 / 5^5 b^9)^7 x (5^11 b^2 / 6^6 a^3)^3
=(6^21a^35/5^35b^63)(5^33b^6/6^18a^9)
=(6³a^26)/ (5²b^57)