Question

hello I want to know that x³-1/x³=(x-1/x)(x²+1/x²+1)​

Answer 1

Answer:

I love this question!

First let's break the equation.

(X+1) (x^2+1) (x^3+1) (x^4+1) >0

And

(X+1) (x^2+1) (x^3+1) (x^4+1) =0

The inequality equation is easy to solve.

Coming to the second equation, atleast any one of the terms should be zero.

So x+1=0

X=-1 (negative sign shows opposite direction to +ve axis, still in same dimension)

And/or

X^2+1=0

X=√-1 = i

(iota means it cannot be shown in the existing number line , so another dimension needed. In 2D, 4 rotations needed to come back to positive, quadrant, you know)

And/or

X^3 +1=0

X=-1 as usual

And /or

X^4+1=0

X= (-1)^(1/4) = √ i

Square root of iota?

Edit

Earlier I wrote that √i calls for another dimension. I was wrong. Square root means make two similar movements to obtain that number. We can rotate two times to get iota in two dimensional plane itself.

Answer 2

Step-by-step explanation:

formula : a^3-b^3 = (a-b)(a^2+ab+b^2)

Ans: thanks for your question