Question

How to find the zero's of `f(x)=1-x^3`?

Answer 1

`x=1`

Explanation:

Using the difference of cubes formula

`a^3-b^3 = (a-b)(a^2+ab+b^2)`, we have

`1-x^3 = 1^3 - x^3`

`=(1-x)(1^2+1*x+x^2)`

`=(1-x)(x^2+x+1)`

Now, setting this equal to `0`, we have

`(1-x)(x^2+x+1) = 0`

`=> 1-x = 0 or x^2+x+1 = 0`

The first equation gives us `x=1` as a solution.

The second equation has no real solutions. We can see this by observing that the discriminant `1^2-4(1)(1) = -3` is negative, meaning the quadratic formula will give us two complex solutions:

`x^2 + x + 1 = 0`

`=> x = (-1+-sqrt(1^2-4(1)(1)))/(2(1))`

`=-1/2+-sqrt(-3)/2`

Thus the only solution to `f(x) = 0` is `x=1`.

Answer 2

I propose a different method.

Solve for `x^3`:

`0 = 1- x^3`

`x^3 = 1`

Get rid of the cube:

`(x^3)^(1/3) = 1^(1/3)`

`x = 1`

Hence, `f(x) = 1 - x^3` has its zero at `x= 1`.

Hopefully this helps!