How do you simplify $\frac{(3ab)^2(4a^3b^4)^3}{(6a^2b)^4}$ ?

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Answer 1 · 2014-11-02T15:14:47

${(3ab)^2(4a^3b^4)^3}/{(6a^2b)^4}$

since $4=2^2$ and $6=3cdot2$ ,

$={(3ab)^2(2^2a^3b^4)^3}/{(3cdot2a^2b)^4}$

by distributing the powers,

$={3^2a^2b^2 cdot 2^6a^9b^{12}}/{3^4 2^4a^8b^4}$

by collecting like factors,

$={3^2 2^6a^{11}b^{14}}/{3^4 2^4a^8b^4}$

By cancelling out $3^2$ , $2^4$ , $a^8$ , and $b^4$ ,

$={2^2a^3b^{10}}/{3}=4/3a^3b^{10}$

I hope that this was helpful.

How do you simplify \frac{(3ab)^2(4a^3b^4)^3}{(6a^2b)^4}\frac{(3ab)^2(4a^3b^4)^3}{(6a^2b)^4}?