Posted by Ari on Thursday, August 23, 2012 at 8:15pm.
(54x^17)^(1/3)
= (27x^15)^(1/3) (2x^2)^(1/3)
= 3x^5 (2^(1/3) x^(2/3)
First I factor 54.
5 4
| |
9 6
|\ \ \
3 3 3 2
{3 3 3} group of threes that can be out of cubed root
Because there are three threes I take them out of the cubed root (# = im going to consider this as my cubed root button) turning #54x^17 into 3#2x^17
Now we factor x^17:
there are 5 groups of 3s in 17 so we would take those out leaving x^2 in the root.
3#2x^17 turns into 3x^5 # 2x^2
Then this simplified is therefore
3x^5 cube root of 2x^2
First I factor 54.
5 4
| |
9 6
|\ \ \
3 3 3 2
{3 3 3} group of threes that can be out of cubed root
Because there are three threes I take them out of the cubed root (# = im going to consider this as my cubed root button) turning #54x^17 into 3#2x^17
Now we factor x^17:
there are 5 groups of 3s in 17 so we would take those out leaving x^2 in the root.
3#2x^17 turns into 3x^5 # 2x^2
Then this simplified is therefore
3x^5 cube root of 2x^2
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