Question

Consider the equation $x^3-1=0$.one of the solutions to this equation is $1$, the other solution $\left(s\right)$ is/are

• A. $-\frac{1}{2}+\frac{\sqrt{3}}{2}i$
• B. $i$
• C. $-i$
• D. $+\frac{1}{2}-\frac{\sqrt{3}}{2}i$

Answer 1

The correct option is A $-\frac{1}{2}+\frac{\sqrt{3}}{2}i$

$x^3-1=0$

$\Rightarrow$ $x^3-1^3=0$

We know,

$(a^3-b^3)=(a-b)(a^2+ab+b^2)$

$\Rightarrow$ $(x-1)(x^2+x+1)=0$

$\Rightarrow$ $x-1=0$ and $x^2+x+1=0$

$\Rightarrow$ $x=1$ [ It is given ]

Now,

$x^2+x+1=0$

Here, $a=1,b=1,c=1$

$\Rightarrow$ $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\Rightarrow$ $x=\frac{-1\pm \sqrt{1^2-4\left(1\right)\left(1\right)}}{2\left(1\right)}$

$\Rightarrow$ $x=\frac{-1\pm \sqrt{1-4}}{2}$

$\Rightarrow$ $x=\frac{-1\pm \sqrt{-3}}{2}$

$\Rightarrow$ $x=\frac{-1\pm \sqrt{3}i}{2}$ [ Since, $\sqrt{-1}=i$ ]

$\Rightarrow$ $x=\frac{-1+\sqrt{3}i}{2}$ and $x=\frac{-1-\sqrt{3}i}{2}$