Question

How do you find the real or imaginary solutions of the equation `x^3+64=0`?

Answer 1

Use the sum of cubes identity to find the Real zero and a quadratic to solve and find the zeros are:

`4" "` and `" "2 +-2sqrt(3)i`

Explanation:

The difference of squares identity can be written:

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

We use this later with `a=(x-2)` and `b=sqrt(-12) = 2sqrt(3)i`, but first...

The sum of cubes identity can be written:

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

Note that `x^3` and `64 = 4^3` are both perfect squares, so the sub of cubes identity applies directly:

`x^3+64 = x^3+4^3`

`color(white)(x^3+64) = (x+4)(x^2-4x+16)`

We can factor the quadratic by completing the square:

`x^2-4x+16 = x^2-4x+4+12`

`color(white)(x^2-4x+16) = (x-2)^2-(2sqrt(3)i)^2`

`color(white)(x^2-4x+16) = ((x-2)-2sqrt(3)i)((x-2)+2sqrt(3)i)`

`color(white)(x^2-4x+16) = (x-2-2sqrt(3)i)(x-2+2sqrt(3)i)`

Hence the zeros of `x^3+64` are:

`4" "` and `" "2+-2sqrt(3)i`