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.