Factor to find all x-intercepts of the function
f(x) = x^6 - x^3
To find the x-intercepts of the function f(x) = x^6 - x^3, we set f(x) equal to zero and solve for x.
So, x^6 - x^3 = 0
We can factor out x^3 from both terms:
x^3(x^3 - 1) = 0
Now, we set each factor equal to zero:
x^3 = 0
This gives us one x-intercept at x = 0.
x^3 - 1 = 0
Using the difference of cubes formula, we can factor this further:
(x - 1)(x^2 + x + 1) = 0
Setting each factor equal to zero gives:
x - 1 = 0
which gives us an x-intercept at x = 1, and
x^2 + x + 1 = 0
This quadratic equation does not factor nicely, so we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
In this case, a = 1, b = 1, and c = 1. Plugging these values into the quadratic formula, we get:
x = (-1 ± √(1 - 4(1)(1)))/(2(1))
x = (-1 ± √(-3))/2
Since the discriminant (-3) is negative, there are no real solutions for this quadratic equation. Therefore, there are no additional x-intercepts for the function f(x) = x^6 - x^3.
In summary, the x-intercepts of the function f(x) = x^6 - x^3 are x = 0 and x = 1.