Will give medal and become fan (accepting only correct answers)! Solve for x. (x^4 - 1) / (x^3) = 0

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Will give medal and become fan (accepting only correct answers)! Solve for x. (x^4 - 1) / (x^3) = 0

Mathematics
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Please explain how you got the answer.
multply both sides by x^3 factor (x^2-1)(x^2+1) and again (x^2-1) = (x+1)(x-1) set to zero x+1=0 x-1=0 What are your roots??? (x^2+1) can never be 0 when y=0 x is?? see graph
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okay so move x^3 to the right side as ~zpupster said multiply both sides by x^3

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Other answers:

Okay so then it would be \[x ^{3} \times (x ^{4} - 1) \div x ^{3} = 0\]
But the x^3 can be cancelled so... the answer would be:\[x ^{4} - 1 = 0\]
Right?
yep you need to multiply *both sides* so 0 times x^3 = 0 now apply difference of square rule
\[\huge\rm a^2-b^2 =(a+b)(a-b)\]
\[(x ^{2} + 1)(x ^{2} - 1)\]
\[(x ^{2}+1)(x+1)(x-1)\]
Is that it?
yep right you would apply difference when there is negative sign so x^2-1 apply difference of square
oopps you already know
yes right set all 3 parentheses equal to zero solve for x
How do you find the square root of -1?
well do you need for this question ?
ohh gotcha so you will get an imaginary solution
Is it going to be \[\pm 1\]
\[\huge rm \sqrt{-1} = i\]
Oh like that! So the answer is x = 1, -1, and i
\[\sqrt{-1}=i \]
yep
Alright thanks again!
my pleasure :=)

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