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

Nov 15, 2016

$x = 1$

#### Explanation:

Using the difference of cubes formula

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$, we have

$1 - {x}^{3} = {1}^{3} - {x}^{3}$

$= \left(1 - x\right) \left({1}^{2} + 1 \cdot x + {x}^{2}\right)$

$= \left(1 - x\right) \left({x}^{2} + x + 1\right)$

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

$\left(1 - x\right) \left({x}^{2} + x + 1\right) = 0$

$\implies 1 - x = 0 \mathmr{and} {x}^{2} + x + 1 = 0$

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 ${1}^{2} - 4 \left(1\right) \left(1\right) = - 3$ is negative, meaning the quadratic formula will give us two complex solutions:

${x}^{2} + x + 1 = 0$

$\implies x = \frac{- 1 \pm \sqrt{{1}^{2} - 4 \left(1\right) \left(1\right)}}{2 \left(1\right)}$

$= - \frac{1}{2} \pm \frac{\sqrt{- 3}}{2}$

Thus the only solution to $f \left(x\right) = 0$ 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:

${\left({x}^{3}\right)}^{\frac{1}{3}} = {1}^{\frac{1}{3}}$

$x = 1$

Hence, $f \left(x\right) = 1 - {x}^{3}$ has its zero at $x = 1$.

Hopefully this helps!