Question

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

Answer 1

"Possible" rational zeros:

`+-1, +-2, +-4, +-8, +-16, +-32, +-64`

Actual zeros:

`+-2`, `-1+-sqrt(3)i`, `1+-sqrt(3)i`

Explanation:

By the rational roots theorem, any rational zeros of `f(x) = x^6-64` are expressible in the form `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:

`+-1, +-2, +-4, +-8, +-16, +-32, +-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=(a-b)(a+b)`

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

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

So:

`x^6-64 = ((x^3)^2-8^2)`

`color(white)(x^6-64) = (x^3-8)(x^3+8)`

`color(white)(x^6-64) = (x^3-2^3)(x^3+2^3)`

`color(white)(x^6-64) = (x-2)(x^2+2x+4)(x+2)(x^2-2x+4)`

Hence rational zeros:

`x = +-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+2x+4 = x^2+2x+1+3`

`color(white)(x^2+2x+4) = (x+1)^2-(sqrt(3)i)^2`

`color(white)(x^2+2x+4) = ((x+1)-sqrt(3)i)((x+1)+sqrt(3)i)`

`color(white)(x^2+2x+4) = (x+1-sqrt(3)i)(x+1+sqrt(3)i)`

Similarly:

`x^2-2x+4 = (x-1-sqrt(3)i)(x-1+sqrt(3)i)`

Hence Complex zeros:

`x = -1+-sqrt(3)i" "` and `" "x = 1+-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]}

Answer 2

`+-2`. See graph

Explanation:

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

We can sort out rational zeros, from real zeros.

Use `(a^3-b^3)=(a-b)(a^2+ab+b^2)`

Any `x^(2N) - a^(2N)=0` has exactly two real roots `x = +-a`,

Here, `(x^6-64)=((x^2)^3-4^3)=(x^2-4)(x^4+4x^2+16)`

So, the zeros are given by

`x^2-4=0 to x = +-2` and

`(x^2)^2+4x^2+16=0 to x^2=-(2+-i2sqrt3)=4e^(+-i2/3pi)`

`x=(4e^(+-i2/3pi))^(1/2)=+-1+-isqrt3`,

upon simplification, using De Moivre;e theorem

So, the real ( and , of course, rational ) zeros are `+-2`.

The complex zeros ( occurring in conjugate pairs ) are

`=+-1+-isqrt3`

Direcly, all-inclusive zeros can be listed as

`x= 2e^(ik/3pi), k = 0, 1, 2, 3, 4, 5.`.

The real roots `+-2` appear for k = 0 and k = 3..