SOLUTION: a problem on a take home test cube root of 125-xcubed =5-x do i first cube both sides to get rid of the radical or should i solve the radical on the left side if i cube both sides

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Question 91637: a problem on a take home test cube root of 125-xcubed =5-x do i first cube both sides to get rid of the radical or should i solve the radical on the left side if i cube both sides i get 125- 5x to the 3 power. if take the cube root of the left side i get 5i times square root of 5 =5-x ihave no idea what to do please help thank you very much!
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
As I understand your problem it is:
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root%283%2C125%29-x%5E3+=+5+-+x
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Recognize that root%283%2C125%29 is just a number that you can get on a calculator.
In this case, the answer can be done in your head. Ask yourself, "What number multiplied
by itself three times results in 125?" It can't be 4*4*4 because that results in 64. How
about 5*5*5. That answer is 125. (Note that -5*-5*-5 will not work because it results in
minus 125.) So you can replace root%283%2C125%29 with 5. When you do that the problem
becomes:
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5+-+x%5E3+=+5+-+x
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Notice that you have 5 on both sides of the equation. You can get rid of them by subtracting
5 from both sides and the equation reduces to:
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-x%5E3+=+-x
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Next add x^3 to both sides of the equation. When you do the -x^3 on the left side disappears
and you are left with:
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0+=+x%5E3+-+x
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Let's transpose sides, just to make this into a little more familiar form:
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x%5E3+-+x+=+0
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Since both terms contain x, let's factor an x out and the problem becomes:
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x%28x%5E2+-1%29=+0
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Then notice that x%5E2+-+1 is the difference between two squares. This is in the standard
factoring form a%5E2+-+b%5E2+=+%28a-b%29%2A%28a%2Bb%29 Applying this rule to x%5E2+-1 we find that
it factors to %28x-1%29%2A%28x%2B1%29 and we can put this into the equation in place of x%5E2-1.
With that substitution the equation we have becomes:
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x%28x-1%29%28x%2B1%29+=+0
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Notice that this equation will be true if any of the factors is equal to zero because
multiplying by a zero on the left side will result in the equation becoming 0 = 0. So
set each of the factors equal to zero and solve for x:
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x+=+0 <=== that's one answer
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x+-+1+=+0+ Solve by adding 1 to both sides to get x+=+1 & that's another answer.
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x+%2B1+=+0+ Solve by subtracting 1 from both sides to get x+=+-1 & that's the last answer.
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In summary, the three answers are x = 0, x = +1, and x = -1.
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You can substitute these answers back into the original problem and see that they each
work.
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Hope this helps you to understand the problem. You didn't have to worry about the cube
root because in this case it was just the number 5. If the cube root had involved a
term containing x, that would have been a whole different problem.
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