simplying indices

$\displaystyle (4y^3)^3 \div 2y^3 $

$\displaystyle 4y^9 \div 2y^3 $

$\displaystyle 4 \div 2 = 2 $ $\displaystyle y^9 \div y^3 = y^6 $

$\displaystyle 2y^6 $

is this working right? Because my book says the correct answer is $\displaystyle 32y^6 $ and I dont understand were 32 came from ?

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http://www.mathhelpforum.com/math-he...389952d9-1.gif

http://www.mathhelpforum.com/math-he...59e50af8-1.gif

http://www.mathhelpforum.com/math-he...214b540c-1.gif http://www.mathhelpforum.com/math-he...65ebae49-1.gif

http://www.mathhelpforum.com/math-he...4aab06fe-1.gif

is this working right? Because my book says the correct answer is http://www.mathhelpforum.com/math-he...90d25bd4-1.gif and I dont understand were 32 came from ?

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hi
you forgot to make 4^3=64 also(Wink)

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Originally Posted by Tweety

$\displaystyle (4y^3)^3 \div 2y^3 $

$\displaystyle 4y^9 \div 2y^3 $

$\displaystyle 4 \div 2 = 2 $ $\displaystyle y^9 \div y^3 = y^6 $

$\displaystyle 2y^6 $

is this working right? Because my book says the correct answer is $\displaystyle 32y^6 $ and I dont understand were 32 came from ?

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Note that $\displaystyle \left(4y^3\right)^3=4^3y^3=\dots$

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Originally Posted by ursa

hi
you forgot to make 4^3=64 also(Wink)

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Hi thanks for answering but I was just wondering why in this expression $\displaystyle (3x^2)^3 \div x^4 $ the number 3 is not 3^3 , ?

as the correct answer for this is $\displaystyle 3x^2 $ so how comes 4 has to be raised to the third power in the expression above?

the answer is wrong as it should be 3^3.