what is the cubed root of x^6???? {x to the sixth exponet}?

2008-06-24 3:50 pm
is it x squared or x cubed?

回答 (1)

2008-06-24 3:58 pm
✔ 最佳答案
x^6 Cubed Root

To find the cubed root, you need to multiply the exponent by 1/3. x^6*1/3 is equal to x^2
2008-06-24 3:55 pm
easiest way to view this as follows:

x*x*x*x*x*x

since cube roots are say 3*3*3 = 27 the cube root of 27 is 3

therefore 3^3 cube root is 3

thus you have x*x or x^2 as your answer since x*x*x cube root is just x.
2008-06-24 3:54 pm
Taking the cubed root is the same thing as raising the number or function to the one third power.

(x^6)^(1/3)
x^2 is the answer
2008-06-24 3:55 pm
Hi,

³√x^6 = x² because you divide the index number on the front of the radical into each exponent. Thew quotient gives the exponent for outside the radical and the remainder, if any, tells the exponent that is still on the variable inside the radical.

x² <== ANSWER

I hope that helps!! :-)
2008-06-24 3:56 pm
³√x^6
= ³√(x^2 * x^2 * x^2)
= x^2
2008-06-24 3:56 pm
it is x square
x^(6)(1/3) = x^2
2008-06-24 3:54 pm
x^2, since (x^2)^3 = x^6
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Ideas: If the cubed root of x^6 is y, then y^3 = x^6.
2016-04-06 10:21 am
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Cube root of 3 = 3^(1/3) Square of Cube root of 3 = {3^(1/3)}^2 = 3^(2/3) .........(1) Square root of 3 = 3^(1/2) Cube of Square Root of 3 = {3^(1/2)}^3 = 3^(3/2) ........(2) Thirteenth Power of 3 = 3^(13) Sixth root of Thirteenth Power of 3 ={3^(13)}^(1/6) = 3^(13/6) .........(3) Product of (1) and (2) = {3^(2/3)} * {3^(3/2)} = 3^{(2/3) + (3/2} = 3^(13/6) ......(4) (No. 4) divided by (No.3) = 1 The result is 1
2008-06-24 7:02 pm
(x^6)^(1/3) = x²
2008-06-24 4:01 pm
the easiest way to look at that is to change from a radical to an exponential.
³√ x^6 = (x^6)^ ( 1/3)
any time you raise a power by a power you multiply the exponents. So you will multiply 6(1/3). This will give you 2.
so:
(x^6)^(1/3) = x²
by changing the radical to a rational exponent made the problem easier for me to see the correct answer. I just need to be sure on how to change from radical to exponential.


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