Question

Find the complex roots of x^3+1=0

Answer 1

`\text{Use:}\ a^3-b^3=(a+b)(a^2-ab+b^2)`

`x^3+1=0\\\\x^3+1^3=0\\\\(x+1)(x^2-(x)(1)+1^2)=0\\\\(x+1)(x^2-x+1)=0\iff x+1=0\ \vee\ x^2-x+1=0\\\\x+1=0\to x=-1\in\mathbb{R}-real\ solution\\\\x^2-x+1=0\\\\x^2-2\cdot x\cdot0.5+1=0\\\\\underbrace{x^2-2\cdot x\cdot0.5+0.5^2}_((a-b)=a^2-2ab+b^2)-0.5^2+1=0\\\\(x-0.5)^2-0.25+1=0\\\\(x-0.5)^2+0.75=0\ \ \ |-0.75\\\\(x-0.5)^2=-0.75\iff x-0.5=\pm√(-0.75)\ \ \ |+0.5\\\\x=0.5-i√(0.75)\ \vee\ x=0.5+i√(0.75)\\\\x=(1)/(2)-i\sqrt{(3)/(4)}\ \vee\ x=(1)/(2)+i\sqrt{(3)/(4)}`

`\boxed{x=(1-i\sqrt3)/(2)\ \vee\ x=(1+i\sqrt3)/(2)}`

`\text{in standard form:}\ \boxed{x=(1)/(2)-(\sqrt3)/(2)i\ \vee\ x=(1)/(2)+(\sqrt3)/(2)i}`