Question 769183
<pre>

x³ - 3 = 0

    x³ = 3

So the zeros are the 3 cube roots of 3

3 = 3[cos(2pn) + i·sin(2pn)]

To get the three cube roots of 3, we use DeMoivre's theorem
using the power of {{{1/3}}}

{{{matrix(2,1,
"",(3(cos(2pi*n) + i*sin(2pi*n)))^(1/3))}}} {{{""=""}}} {{{matrix(2,1,
"",3^(1/3)(cos(2pi*n) + i*sin(2pi*n))^(1/3))}}}

&#8731;<span style="text-decoration: overline">3</span>[cos{{{(expr(2pi/3)*n)}}} + i·sin{{{(expr(2pi/3)*n)}}}]

To get the first cube root, we let n=0

&#8731;<span style="text-decoration: overline">3</span>[cos{{{(expr(2pi/3)*0)}}} + i·sin{{{(expr(2pi/3)*0)}}}] = &#8731;<span style="text-decoration: overline">3</span>[cos(0) + i·sin(0)] = &#8731;<span style="text-decoration: overline">3</span>[1 + 0i] = &#8731;<span style="text-decoration: overline">3</span>

To get the second cube root of 3, we let n=1

&#8731;<span style="text-decoration: overline">3</span>[cos{{{(expr(2pi/3)*1)}}} + i·sin{{{(expr(2pi/3)*1)}}}] = &#8731;<span style="text-decoration: overline">3</span>[cos{{{((2pi)/3)}}} + i·sin{{{((2pi)/3)}}}] = &#8731;<span style="text-decoration: overline">3</span>[{{{-1/2}}} + i·{{{sqrt(3)/2}}}] =

{{{-root(3,3)/2}}}{{{""+""}}}{{{i*expr( (root(3,3)root("`",3))/2)}}} 

and we can write 

{{{root(3,3)root("`",3)}}}{{{""=""}}}{{{matrix(2,1,"",3^(1/3)*3^(1/2))}}}{{{""=""}}}{{{matrix(2,1,"",3^(2/6)*3^(3/6))}}}{{{""=""}}}{{{matrix(2,1,"",3^(5/6))}}}{{{""=""}}}{{{root(6,3^5)}}}{{{""=""}}}{{{root(6,243)}}}

So the second cube root of 3 is {{{-root(3,3)/2}}}{{{""+""}}}{{{expr(root(6,243)/2)*i}}}

To get the third cube root of 3, we let n=2

&#8731;<span style="text-decoration: overline">3</span>[cos{{{(expr(2pi/3)*2)}}} + i·sin{{{(expr(2pi/3)*2)}}}] = &#8731;<span style="text-decoration: overline">3</span>[cos{{{((4pi)/3)}}} + i·sin{{{((4pi)/3)}}}]

and since cos{{{((4pi)/3)}}} = cos{{{((2pi)/3)}}}, and sin{{{((4pi)/3)}}} = -sin{{{((2pi)/3)}}},
 
the third cube root is just the conjugate of the 
second cube root, or

{{{-root(3,3)/2}}}{{{""-""}}}{{{expr(root(6,243)/2)*i}}}

So the zeros of x³ - 3 are:

&#8731;<span style="text-decoration: overline">3</span>, {{{-root(3,3)/2}}}{{{""+""}}}{{{expr(root(6,243)/2)*i}}}, and {{{-root(3,3)/2}}}{{{""-""}}}{{{expr(root(6,243)/2)*i}}}

Edwin</pre>