HELP!

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thetomps

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Can someone explain how to do this problem. It is for my sister. She needs help. I want to be able to show her the steps how to do it. Any help is greatly appreciated it. Thanks.

Kay

The solution to the equation x^3-8=0 are all the cube roots of 8.
a.) How many cube roots of 8 are there?
b.) 2 is obviously a cube root of 8; find all others

I'm sorry I have been out of school for awhile so I don't remember this stuff.
 
I'm guessing you don't REALLY want to know the other two. Technically, EVERY Real Number has THREE Cube Roots. The trouble is, the three Cube Roots are not necessarily Real Numbers. Since this is posted in "Beginning Algebra", I am somewhat reticent to say much mroe about it. If you haven't gotten to factorign a difference of cubes or the qquadratic formula, the answer definitely will be a problem.
 
I'm trying to explain this to my sis. She is a freshman in college. She has a class tonight to turn it into. She doesn't know how to do it. She thought she knew what she was doing but she has been in tears all day that she is stuck. I wanted to show her a break down on how to do it. Can you help me?
 
thetomps said:
The solution to the equation x^3-8=0 are all the cube roots of 8.
a.) How many cube roots of 8 are there?
Click to expand...
For any number, if one includes complex values, there are n n-th roots. For any n-degree polynomial, if once includes complex values, there are n roots.

thetomps said:
b.) 2 is obviously a cube root of 8; find all others
Click to expand...
Apply the difference-of-cubes formula to factor x<sup>3</sup> - 8 into a linear factor and a quadratic factor. Then apply the Quadratic Formula to find the remaining two roots.

Hope that helps a bit. If your sister gets stuck, please have her reply, showing all the steps she has attempted. Thank you.

Eliz.
 
thetomps said:
b.) x^3 - 8
(x+4) (x-2) x
Click to expand...
I'm sorry, but I don't understand. How does "(x + 4)(x - 2)(x) = x<sup>3</sup> + 2x<sup>2</sup> - 8x" relate to the exercise at hand?

Thank you.

Eliz.
 
Staple,
I am lost. Please help. This is the only problem like this I have on my homework. I have gone thru my book and it doesnt help either.

Abby

:(
 
Ok, no more second guessing.
x^3-8 =
(x-2)(x^2+2x+4)
Use the quadratic equation on the second term
x=(-2+sqrt(2²-4*1*4))/2 =
-1+sqrt(-3)
The answers are 2,-1+sqrt(-3),-1-sqrt(-3)
 
If they're asking you to solve the difference of cubes, they must have covered the formula for factoring the difference of cubes. Just apply that, which will give you a linear and a quadratic factor. To the quadratic factor, apply the Quadratic Formula. This will give you the other two solutions.

In other words, this exercise may be a little different, but it's only really asking that you apply the formulas you've memorized.

Eliz.
 
so what should i put for the answer for a and b. Thanks again for helping me and my sis.

Abby
:'(
 
How many cube roots of 8 are there? Three.

2 is obviously a cube root of 8; find all others
2cos(2[pi]/3)+2isin(2[pi]/3) and 2cos(−2[pi]/3)+2isin(−2[pi]/3)
 
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