Question 10848
1) {{{x^3 - 8 = 0 }}} Add 8 to both sides, which you did already.

{{{x^3 = 8}}} Take the cube root of both sides.

{{{x = 2}}} 

  -2 is not a root since: {{{ (-2)^3 = (-2)(-2)(-2)}}} = -8

2) {{{x^3 + 27 = 0}}} Subtract 27 from both sides.

{{{x^3 = -27}}} Take the cube root of both sides.

{{{x = -3}}} 

+3 is not a root since: {{{(+3)^3 = (+3)(+3)(+3)}}} = +27

Please ignore the above...it's wrong!

1)  {{{x^3 - 8 = 0}}}  This is the difference of two cubes which can be factored as follows:

{{{(x - 2)(x^2 + 2x + 4) = 0}}}

{{{x - 2 = 0}}}, so {{{x = 2}}} One root.

{{{x^2 + 2x  + 4 = 0}}} Use the quadratic formula: {{{x = (-b+-sqrt(b^2 - 4ac))/2a}}}

{{{x = (-2+-sqrt(4-16))/2}}}

{{{x = (-2+-sqrt(-12))/2}}}

{{{x = -1+-sqrt(-3*4)/2}}}

{{{x = -1+sqrt(3)i}}} or {{{x = -1-sqrt(3)i}}} The other two roots.

2) {{{x^3 + 27 = 0}}} This is the sum of two cubes which can be factored as follows:

{{{(x + 3)(x^2 - 3x + 9) = 0}}} Setting each factor = 0.
{{{(x + 3) = 0}}}
{{{x = -3}}} This is one of the roots.

{{{x^2 - 3x + 9 = 0}}} Use the quadratic formula: {{{x=(-b+-sqrt(b^2-4ac))/2a}}}

{{{x=(3+-sqrt(9-36))/2}}}
{{{x=(3/2)+-sqrt(-27)/2}}}
{{{x=(3/2)+3sqrt(3)i/2}}} or {{{x=(3/2)-3sqrt(3)i/2}}} The other two roots.

These can also be written:

{{{x = (3/2)(1+sqrt(3)i)}}} and
{{{x = (3/2)(1-sqrt(3)i)}}}