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

2 Answers
Nov 15, 2016

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.

Nov 15, 2016

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!