NIMCET 2011 — Mathematics PYQ

The equation is $x^2-2x+4=0$.

The roots are given by the quadratic formula:

2. Relate to Complex Numbers (Cube Roots of Unity):

Notice that the roots can be written in polar form:

$1+\sqrt{3}i=2(\frac{1}{2}+\frac{\sqrt{3}}{2}i)=2(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3})=2e^{i\pi /3}$

Similarly, $1-\sqrt{3}i=2e^{-i\pi /3}$.

Alternatively, we can use the roots of $x^3+8=0$, which is $(x+2)(x^2-2x+4)=0$.

This means $\alpha$ and $\beta$ satisfy $x^3=-8$.

3. Calculate ${\alpha }^6$ and ${\beta }^6$:

If $\alpha$ is a root, then ${\alpha }^3=-8$.

Squaring both sides:

Similarly, since $\beta$ is also a root of the same quadratic:

4. Find the final sum:

Conclusion:

The value of ${\alpha }^6+{\beta }^6$ is $128$.

The correct option is (b).