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

Question

anonymous · Answer

Please explain how you got the answer.

zpupster · 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

nnesha · Answer

okay so move x^3 to the right side 
as ~zpupster said multiply both sides by x^3

anonymous · Answer

Okay so then it would be $$x ^{3} \times (x ^{4} - 1) \div x ^{3} = 0$$

anonymous · Answer

But the x^3 can be cancelled so... the answer would be:$$x ^{4} - 1 = 0$$

anonymous · Answer

Right?

nnesha · Answer

yep you need to multiply *both sides* so 0 times x^3 = 0
now apply difference of square rule

nnesha · Answer

$$\huge\rm a^2-b^2 =(a+b)(a-b)$$

anonymous · Answer

$$(x ^{2} + 1)(x ^{2} - 1)$$

anonymous · Answer

$$(x ^{2}+1)(x+1)(x-1)$$

anonymous · Answer

Is that it?

nnesha · Answer

yep right 
you would apply difference when there is negative sign 
so x^2-1 apply difference of square

nnesha · Answer

oopps you already know

nnesha · Answer

yes right set all 3 parentheses equal to zero solve for x

anonymous · Answer

How do you find the square root of -1?

nnesha · Answer

well do you need for this question ?

nnesha · Answer

ohh gotcha 
so you will get an imaginary solution

anonymous · Answer

Is it going to be $$\pm 1$$

nnesha · Answer

$$\huge
rm \sqrt{-1} = i$$

anonymous · Answer

Oh like that! So the answer is x = 1, -1, and i

nnesha · Answer

$$\sqrt{-1}=i $$

nnesha · Answer

yep

anonymous · Answer

Alright thanks again!

nnesha · Answer

my pleasure :=)