Question

8x^3 = 1

Answer 1

there are three zeros.
one is as antinode calculated and there are 2 others.
\[8x^3=1,~or~8x^3-1=0,\left( 2x \right)^3-1^3=0\]
\[\left( 2x-1 \right)\left( 4x^2+2x+1 \right)=0\]
\[\left[ a^3-b^3=\left( a-b \right)\left( a^2+ab+b^2 \right) \right]\]
either 2x-1=0,x=1/2
or \[4x^2+2x+1=0\]
\[x=\frac{ -2\pm \sqrt{2^2-4*4*1} }{2*4 }=\frac{ -2\pm \sqrt{-12} }{ 8 }\]
\[=\frac{ -2\pm2\sqrt{-3} }{ 8 }=\frac{ -1\pm \iota \sqrt{3} }{ 4 }\]
these two zeros are complex.