Question

If the cube roots of unity are 1, $\omega ,{\omega }^2$, then the roots of the equation $(x-1{)}^3+8=0$, are

• A. $-1,-1,1-2{\omega }^2$
• B. $-1,-1+2\omega ,-1-2{\omega }^2$
• C. $-1,1+2\omega ,1+2{\omega }^2$
• D. $-1,1-2\omega ,1-2{\omega }^2$

Answer 1

The correct option is D $-1,1-2\omega ,1-2{\omega }^2$

$Method\left(1\right)$ : By making the equation from the given roots)

Let us consider $x=-1,-1,-1$

$\therefore$ Required equation from given roots is

$(x+1)(x+1)(x+1)=0$

$(x+1{)}^3=0$ which does not match with the given equation

$(x-1{)}^3+8=0$ so$x=-1,-1,-1$ cannot be the proper choice. Again consider $x=-1,-1,+2\omega ,-1-2{\omega }^2$

$\therefore$ Required equation from given roots is

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

$\Rightarrow (x+1)\left[\right(x+1{)}^2+(x+1)(2{\omega }^2-2\omega )-4{\omega }^3]=0$

$\Rightarrow (x+1)\left[\right(x+1{)}^2+2(x+1)({\omega }^2-\omega )-4]=0$

$\Rightarrow (x+1{)}^3+2(x+1{)}^2({\omega }^2-\omega )-4(x+1)=0$

Which cannot be expressed in the form of given equation $(x+1{)}^3+8=0$

Now consider the roots

$x_1=-1,x_2=1-2\omega ,x_3=1-2{\omega }^2$
and the equation with these roots is given by

$x^3$ - (sum of the roots) $x^2+x$ (Product of roots taken two at a time) - Product of roots taken all at a time = 0

Now sum of roots $=x_1+x_2+x_3=-1+1-2\omega +1-2{\omega }^2=3$

Product of roots taken two at a time $=-1+2\omega -1+2{\omega }^2+1+2({\omega }^2+\omega )+4{\omega }^3=3$

Product of roots taken all at a time

$=(-1)\left[\right(1-2\omega \left)\right(1-2{\omega }^2\left)\right]=-7$

$\therefore$ Equation formed is $x^3-3x^2+3x+7=0$

$\Rightarrow x^3-3x^2+3x-1+8=0$

$\Rightarrow (x-1{)}^3+8=0$

which matched with given equation.

$\therefore -1,1-2\omega ,1-2{\omega }^2$ are the roots of given equation.

$Method2$ : (by taking cross checking)

As $(x-1{)}^3+8=0$

and $x=-1$ satisfies $(x-1{)}^3+8=0$ i.e. $(-2{)}^3+8=0$

Similarly for $1-2\omega$ we have $(x-1{)}^3+8=0$

$\Rightarrow (1-2\omega -1{)}^3+8=0\Rightarrow (-2\omega {)}^3+8=0\Rightarrow -8+8=0$

and for $1-2{\omega }^2$ we have $(1-2{\omega }^2-1{)}^3+8=0$

$\Rightarrow {\omega }^6(-8)+\left(8\right)=0\Rightarrow 0=0$

$\therefore -1,1-2\omega ,1-2{\omega }^2$ are roots of $(x-1{)}^3+8=0$ but on the other hand the other roots does not satisfies the equation $(x-1{)}^3+8=0$