# simplying indices

• Jan 17th 2009, 08:38 AM
Tweety
simplying indices
$(4y^3)^3 \div 2y^3$

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

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

$2y^6$

is this working right? Because my book says the correct answer is $32y^6$ and I dont understand were 32 came from ?
• Jan 17th 2009, 09:02 AM
ursa
• Jan 17th 2009, 09:04 AM
Chris L T521
Quote:

Originally Posted by Tweety
$(4y^3)^3 \div 2y^3$

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

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

$2y^6$

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

Note that $\left(4y^3\right)^3=4^3y^3=\dots$
• Jan 17th 2009, 09:13 AM
Tweety
Quote:

Originally Posted by ursa
hi
you forgot to make 4^3=64 also(Wink)

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

as the correct answer for this is $3x^2$ so how comes 4 has to be raised to the third power in the expression above?
• Jan 17th 2009, 09:17 AM
hmmmm
the answer is wrong as it should be 3^3.