What are all the possible rational zeros for f(x)=x^6-64?

Dec 20, 2016

Answer:

"Possible" rational zeros:

$\pm 1 , \pm 2 , \pm 4 , \pm 8 , \pm 16 , \pm 32 , \pm 64$

Actual zeros:

$\pm 2$, $- 1 \pm \sqrt{3} i$, $1 \pm \sqrt{3} i$

Explanation:

By the rational roots theorem, any rational zeros of $f \left(x\right) = {x}^{6} - 64$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 64$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only "possible" rational zeros are:

$\pm 1 , \pm 2 , \pm 4 , \pm 8 , \pm 16 , \pm 32 , \pm 64$

In practice we can factor this sextic as a difference of squares, difference of cubes and sum of cubes, allowing us to find the Real zeros and quadratics for the Complex zeros.

We will use the following identities:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

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

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

So:

${x}^{6} - 64 = \left({\left({x}^{3}\right)}^{2} - {8}^{2}\right)$

$\textcolor{w h i t e}{{x}^{6} - 64} = \left({x}^{3} - 8\right) \left({x}^{3} + 8\right)$

$\textcolor{w h i t e}{{x}^{6} - 64} = \left({x}^{3} - {2}^{3}\right) \left({x}^{3} + {2}^{3}\right)$

$\textcolor{w h i t e}{{x}^{6} - 64} = \left(x - 2\right) \left({x}^{2} + 2 x + 4\right) \left(x + 2\right) \left({x}^{2} - 2 x + 4\right)$

Hence rational zeros:

$x = \pm 2$

The remaining quadratics have negative discriminants, so Complex zeros, but we can factor them using the difference of squares identity as follows:

${x}^{2} + 2 x + 4 = {x}^{2} + 2 x + 1 + 3$

$\textcolor{w h i t e}{{x}^{2} + 2 x + 4} = {\left(x + 1\right)}^{2} - {\left(\sqrt{3} i\right)}^{2}$

$\textcolor{w h i t e}{{x}^{2} + 2 x + 4} = \left(\left(x + 1\right) - \sqrt{3} i\right) \left(\left(x + 1\right) + \sqrt{3} i\right)$

$\textcolor{w h i t e}{{x}^{2} + 2 x + 4} = \left(x + 1 - \sqrt{3} i\right) \left(x + 1 + \sqrt{3} i\right)$

Similarly:

${x}^{2} - 2 x + 4 = \left(x - 1 - \sqrt{3} i\right) \left(x - 1 + \sqrt{3} i\right)$

Hence Complex zeros:

$x = - 1 \pm \sqrt{3} i \text{ }$ and $\text{ } x = 1 \pm \sqrt{3} i$

These zeros form the vertices of a regular hexagon in the Complex plane:

graph{((x-2)^2+y^2-0.01)((x+2)^2+y^2-0.01)((x-1)^2+(y-sqrt(3))^2-0.01)((x-1)^2+(y+sqrt(3))^2-0.01)((x+1)^2+(y-sqrt(3))^2-0.01)((x+1)^2+(y+sqrt(3))^2-0.01)(x^2+y^2-4) = 0 [-5, 5, -2.5, 2.5]}

Dec 21, 2016

Answer:

$\pm 2$. See graph

Explanation:

graph{x^6-64 [-3, 3, -500, 500]}

We can sort out rational zeros, from real zeros.

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

Any ${x}^{2 N} - {a}^{2 N} = 0$ has exactly two real roots $x = \pm a$,

Here, $\left({x}^{6} - 64\right) = \left({\left({x}^{2}\right)}^{3} - {4}^{3}\right) = \left({x}^{2} - 4\right) \left({x}^{4} + 4 {x}^{2} + 16\right)$

So, the zeros are given by

${x}^{2} - 4 = 0 \to x = \pm 2$ and

${\left({x}^{2}\right)}^{2} + 4 {x}^{2} + 16 = 0 \to {x}^{2} = - \left(2 \pm i 2 \sqrt{3}\right) = 4 {e}^{\pm i \frac{2}{3} \pi}$

$x = {\left(4 {e}^{\pm i \frac{2}{3} \pi}\right)}^{\frac{1}{2}} = \pm 1 \pm i \sqrt{3}$,

upon simplification, using De Moivre;e theorem

So, the real ( and , of course, rational ) zeros are $\pm 2$.

The complex zeros ( occurring in conjugate pairs ) are

$= \pm 1 \pm i \sqrt{3}$

Direcly, all-inclusive zeros can be listed as

$x = 2 {e}^{i \frac{k}{3} \pi} , k = 0 , 1 , 2 , 3 , 4 , 5.$.

The real roots $\pm 2$ appear for k = 0 and k = 3..