2007-04-27 07:00:43 · 11 answers · asked by ~*¤*~Brit ~ 1 in Science & Mathematics ➔ Mathematics
(8x^2y^6)^(1/3) break it up (note 1/3 = cube root)
=(8)^(1/3) * (x^2)^(1/3) * (y^6)^1/3
=2*(x^2/3)*(y^6/3) when taking someting to a power of a power u multiply the two powers now we get
2(y^2)(x^2/3)
If you need further help or clarification email me or im me. Good Luck.
2007-04-27 07:06:44 · answer #1 · answered by Anonymous · 4⤊ 1⤋
Take the cube root of 8, which is asking what number times itself 3 times is 8 and that is 2
and simply divide the exponents by 3 to get
x^2/3 and y^6/3 to yield
2x^2/3y^2
2007-04-27 14:19:36 · answer #2 · answered by smokedhamm69 1 · 0⤊ 1⤋
cube root of 8x^2y^6=2x^(2/3)y^2
or 2 * cube root of x^2 * Y^2
2007-04-27 14:09:42 · answer #3 · answered by Anonymous · 0⤊ 1⤋
You can write 8 as 2^3
Now you have 2^3 * x^2 * y^6
cube root means raising this to the power 1/3 or dividing all the powers by 3.
we get 2 * x^(2/3) * y^2
2007-04-27 14:08:23 · answer #4 · answered by Dr D 7 · 2⤊ 1⤋
cube root of 8x^2y^6= 2x^(2/3)y^2
2007-04-27 14:06:23 · answer #5 · answered by (^InLove^) 3 · 1⤊ 2⤋
it would be easy if you have a TI-83 calc. anyway, the answer is 2y^2 square root of x^2
2007-04-27 14:10:19 · answer #6 · answered by Ariel 2 · 0⤊ 2⤋
2 * x^(2/3) y^2
2007-04-27 14:07:22 · answer #7 · answered by sandy 2 · 1⤊ 2⤋
(8x^2y^6)^1/3 cube root = ^1/3
(8x^2y*6)^1/3
(8x^12y)^1/3
8^1/3*x^12y^1/3
2x^4y
qed(quite easily done)
2007-04-27 14:09:37 · answer #8 · answered by harry m 6 · 0⤊ 3⤋
I'm not sure of your grouping, but I think it's
2*x^(2/3) * y^2
2007-04-27 14:08:08 · answer #9 · answered by Robert L 7 · 1⤊ 1⤋
=2y^2*x^(2/3)
2007-04-27 14:06:31 · answer #10 · answered by santmann2002 7 · 1⤊ 2⤋