JEE MAIN - Mathematics (2005 - No. 10)
If the cube roots of unity are 1, $$\omega \,,\,{\omega ^2}$$ then the roots of the equation $${(x - 1)^3}$$ + 8 = 0, are :
$$ - 1, - 1 + 2\,\,\omega , - 1 - 2\,\,{\omega ^2}$$
$$ - 1, - 1, - 1$$
$$ - 1,1 - 2\omega ,1 - 2{\omega ^2}$$
$$ - 1,1 + 2\omega ,1 + 2{\omega ^2}$$
Explanation
$${\left( {x - 1} \right)^3} + 8 = 0$$
$$ \Rightarrow \left( {x - 1} \right) = \left( { - 2} \right){\left( 1 \right)^{1/3}}$$
$$ \Rightarrow x - 1 = - 2\,\,\,$$ or $$\,\,\, - 2\omega \,\,\,\,$$ or $$\,\,\,\, - 2{\omega ^2}$$
or $$\,\,\,x = - 1\,\,\,$$ or $$\,\,\,1 - 2\omega \,\,\,$$ or $$\,\,\,1 - 2{\omega ^2}.$$
$$ \Rightarrow \left( {x - 1} \right) = \left( { - 2} \right){\left( 1 \right)^{1/3}}$$
$$ \Rightarrow x - 1 = - 2\,\,\,$$ or $$\,\,\, - 2\omega \,\,\,\,$$ or $$\,\,\,\, - 2{\omega ^2}$$
or $$\,\,\,x = - 1\,\,\,$$ or $$\,\,\,1 - 2\omega \,\,\,$$ or $$\,\,\,1 - 2{\omega ^2}.$$
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