माना alpha तथा beta समीकरण x ^{2}+ x +1=0 के | vedclass

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Question

Roots of the equation ${x^2} + x + 1 = 0$ are $\alpha  = \omega $ and $\beta  = {\omega ^2}$ where $\omega ,{\omega ^2}$

Vedclass · Accepted Answer

$y^3$

Vedclass · Answer

Roots of the equation ${x^2} + x + 1 = 0$ are $\alpha  = \omega $ and $\beta  = {\omega ^2}$ where $\omega ,{\omega ^2}$are complex cube roots of unity

$	herefore \Delta  = \left| {\begin{array}{*{20}{c}}
{y + 1}&\omega &{{\omega ^2}}\
\omega &{y + {\omega ^2}}&1\
{{\omega ^2}}&1&{y + \omega }
\end{array}} ight|$

${R_1} 	o {R_1} + {R_2} + {R_3}$

$ \Rightarrow \Delta  = \left| {\begin{array}{*{20}{c}}
1&1&1\
\omega &{y + {\omega ^2}}&1\
{{\omega ^2}}&1&{y + \omega }
\end{array}} ight|$

Expanding along ${R_1}$, we get

$\Delta  = y.{y^2} \Rightarrow D = {y^3}$

Or

If $\alpha  = {\omega ^2},\beta  = \omega $ we get same value or on expansion using $\alpha  + \beta  =  - 1,\alpha \beta  = 1$ we get value ${y^3}$.

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