How do you solve $x^3 + 64 = 0$ ?

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Answer 1 · 2016-01-21T14:24:08

$x=-4,2+-2sqrt3i$

Notice that this is a sum of cubes, which is factorable as follows:

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

Thus, $x^3+64$ is factorable into

$x^3+4^3=(x+4)(x^2-4x+16)=0$

Now, we have one linear factor and one quadratic factor.

$(x+4)(x^2-4x+16)=0$

We can set each of these equal to $0$ individually to find the values of $x$ that make the whole expression equal $0$ .

$x+4=0" "=>" "x=-4$

$x^2-4x+16=0" "=>" "x=(4+-sqrt(16-64))/2$

$=>x=(4+-4sqrt3i)/2" "=>" "x=2+-2sqrt3i$

These are two imaginary solutions.

How do you solve x^3 + 64 = 0x^3 + 64 = 0?