SOLUTION: 1. x^3-4=2, 2. And is this how you solve this problem? x3/2=125, x3/2=5^3 (3/2) 2/3=(5 3/1) 2/3 which the answer is 5^2 or write just 25 as the answer? 3. And how do I solve a alge

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Question 164837: 1. x^3-4=2, 2. And is this how you solve this problem? x3/2=125, x3/2=5^3 (3/2) 2/3=(5 3/1) 2/3 which the answer is 5^2 or write just 25 as the answer? 3. And how do I solve a algerbracially and check your potential solutions: (x)+20-x=0? 4. And if y=(x)-2 how do I feel in the following table for x=0,1,2?
Answer by MRperkins(300) (Show Source):
You can put this solution on YOUR website!
Good try. I like to see students at least attempt a problem so thank you for your attempt. I hope you are able to understand this problem better after my explanation:
Given:
Solve for x:
Method 1: add 4 to both sides

take the cubed root of both sides
x=2
Method 2: subtract 2 from both sides

factor this problem using the special rule for the difference of two perfect cubes which says:"
or
can further be factored to (x+2)(x+2)
so
x^3-6 when factored equals:(x-2)(x+2)(x+2)
so
set each set equal to 0
x-2=0; x+2=0
so

Check your answers by substituting 2 for x: 2^3 is 2*2*2 or 6;6-4 is 2 and 2=2 is true.
now check -2. (-2)^3 =-2*-2*-2 or -8; -8-4 is -12; -12 does not equal 2 so it is not an answer. Therefore your only answer is x=2.

Part 3 of your question does not make sense: (x)+20-x=0
if you combine like terms then you get 20=0 which is not true. I think I may have answered what you need in Part 1 of my answer.
Part 4
The question is asking you to make a chart of y values when x =0 x=1 and x=2.
So in the equation y=x-2
x:y
0:-2 when x=0 y=-2
1:-1 when x=1 y=-1
2:0 when x=2 y=0
I hope this helps

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