we are given
[tex]27x^3-1=0[/tex]
we can also write it as
[tex](3x)^3-(1)^3=0[/tex]
now, we can use factor formula
[tex]a^3-b^3=(a-b)(a^2+ab+b^2)[/tex]
we can use above formula
we get
[tex](3x)^3-(1)^3=(3x-1)((3x)^2+3x\times 1+(1)^2)[/tex]
now, we can simplify it
[tex]27x^3-1=(3x-1)(9x^2+3x+1)[/tex]
now, we can set it to 0
[tex]27x^3-1=(3x-1)(9x^2+3x+1)=0[/tex]
and then we can solve for x
[tex](3x-1)(9x^2+3x+1)=0[/tex]
[tex]3x-1=0[/tex]
[tex]x=\frac{1}{3}[/tex]
[tex]9x^2+3x+1=0[/tex]
now, we can use quadratic formula
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
now, we can plug values
and we get
[tex]x=\frac{-3\pm \sqrt{3^2-4\cdot \:9\cdot \:1}}{2\cdot \:9}[/tex]
[tex]x=-\frac{1}{6}+i\frac{\sqrt{3}}{6},\:x=-\frac{1}{6}-i\frac{\sqrt{3}}{6}[/tex]
So, we will get solution as
[tex]x=\frac{1}{3},\:x=-\frac{1}{6}+i\frac{\sqrt{3}}{6},\:x=-\frac{1}{6}-i\frac{\sqrt{3}}{6}[/tex]...............Answer
