Question

Check whether the following equation is True or False, ${x^3} - {y^3} = (x - y)(\omega x - {\omega ^2}y)({\omega ^2}x - \omega y)$
(A). True
(B). False

Answer 1

Hint- In this question, just follow the simple approach that uses the identity ${x^3} - {y^3} = (x - y)({x^2} + xy + {y^2})$ . After this just solve the second component $({x^2} + xy + {y^2})$ of the equation using the Quadratic formula to split that into two components and checking whether the given equation is true or false.

Complete step-by-step solution -
We know that, ${x^3} - {y^3} = (x - y)({x^2} + xy + {y^2})$ -------(1)
Considering the component $({x^2} + xy + {y^2})$ we want to simplify and split this one. For that we will first equate that to 0.
$\Rightarrow {x^2} + (y)x + {y^2} = 0$
Now relating this to $a{x^2} + bx + c = 0$
Here, $D = {b^2} - 4ac$ and here we have $a = 1,b = y,c = {y^2}$
$\Rightarrow D = {(y)^2} - 4 \times 1 \times {y^2} = - 3{y^2}$
Since, $D < 0$
The roots are complex.
$\Rightarrow x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$
Substituting the values, we get,
$\Rightarrow x = \dfrac{{ - y \pm \sqrt { - 3{y^2}} }}{2}$
$\Rightarrow x = y\left( {\dfrac{{ - 1 \pm i\sqrt 3 }}{2}} \right)$
Now, we know that $\dfrac{{ - 1 + i\sqrt 3 }}{2} = \omega ,\dfrac{{ - 1 - i\sqrt 3 }}{2} = {\omega ^2}$
So, $x = y\omega$ and $x = y{\omega ^2}$
$\Rightarrow {x^2} + (y)x + {y^2} = 0$ can be rewritten as $(x - y\omega )(x - y{\omega ^2}) = 0$
Now, $(x - y\omega )(x - y{\omega ^2}) = 0$ can be rewritten as ${\omega ^3}(x - y\omega )(x - y{\omega ^2}) = 0$ because ${\omega ^3} = 1$
We can make ${\omega ^3}$ as ${\omega ^2} \times \omega$ such that $\omega (x - y\omega ) \times {\omega ^2}(x - y{\omega ^2}) = 0$
It can be rewritten as $(\omega x - y{\omega ^2})({\omega ^2}x - y{\omega ^4}) = 0$
$\Rightarrow (\omega x - y{\omega ^2})({\omega ^2}x - y{\omega ^3}\omega ) = 0$
Now, ${\omega ^3} = 1$
$\Rightarrow (\omega x - y{\omega ^2})({\omega ^2}x - y\omega ) = 0$
or, $(\omega x - {\omega ^2}y)({\omega ^2}x - \omega y) = 0$
So, $({x^2} + xy + {y^2})$ can be rewritten as $(\omega x - {\omega ^2}y)({\omega ^2}x - \omega y)$
Put this in (1)
$\Rightarrow {x^3} - {y^3} = (x - y)(\omega x - {\omega ^2}y)({\omega ^2}x - \omega y)$
Hence, the given equation is True.
$\therefore$ Option A. True is the correct answer.

Note- In such types of questions, just keep in mind the algebraic identity like ${x^3} - {y^3} = (x - y)({x^2} + xy + {y^2})$ , also keep in mind the quadratic formula to find the roots of the quadratic equation and also remember that for complex roots $\dfrac{{ - 1 + i\sqrt 3 }}{2} = \omega ,\dfrac{{ - 1 - i\sqrt 3 }}{2} = {\omega ^2}$ where $\omega$ stands for cube root of unity.