Let a,b C.

Question

Let $a,b\in C$. Let $\alpha ,\beta $ be the roots of the equation ${x}^{2}+ax+b=0$. If $\beta -\alpha =\sqrt{11}$ and ${\beta }^{2}-{\alpha }^{2}=3i\sqrt{11}$, then ${({\beta }^{3}-{\alpha }^{3})}^{2}$ is equal to:

Accepted Answer

Given:

$\alpha ,\beta $ are roots of ${x}^{2}+ax+b=0$

$\beta -\alpha =\sqrt{11}$

${\beta }^{2}-{\alpha }^{2}=3i\sqrt{11}$

We have to find ${({\beta }^{3}-{\alpha }^{3})}^{2}$.

Now,

${\beta }^{2}-{\alpha }^{2}=(\beta -\alpha )(\beta +\alpha )$

Substituting the given values,

$3i\sqrt{11}=\sqrt{11}(\beta +\alpha )$

Since $\sqrt{11}
e 0$, we get

$\beta +\alpha =3i$

Now use the identity

${\beta }^{3}-{\alpha }^{3}=(\beta -\alpha )({\beta }^{2}+\alpha \beta +{\alpha }^{2})$

So first let us find ${\beta }^{2}+\alpha \beta +{\alpha }^{2}$.

We know

$(\beta +\alpha {)}^{2}={\beta }^{2}+{\alpha }^{2}+2\alpha \beta $

Also,

$\alpha \beta =\frac{(\alpha +\beta {)}^{2}-(\beta -\alpha {)}^{2}}{4}$

Substitute the values:

$\alpha +\beta =3i,\, \beta -\alpha =\sqrt{11}$

Thus,

$\alpha \beta =\frac{(3i{)}^{2}-(\sqrt{11}{)}^{2}}{4}=\frac{-9-11}{4}=\frac{-20}{4}=-5$

Now,

${\alpha }^{2}+{\beta }^{2}=(\alpha +\beta {)}^{2}-2\alpha \beta $

So,

${\alpha }^{2}+{\beta }^{2}=(3i{)}^{2}-2(-5)=-9+10=1$

Hence,

${\beta }^{2}+\alpha \beta +{\alpha }^{2}=1+(-5)=-4$

Therefore,

${\beta }^{3}-{\alpha }^{3}=(\beta -\alpha )({\beta }^{2}+\alpha \beta +{\alpha }^{2})=\sqrt{11}\cdot (-4)=-4\sqrt{11}$

Now square it:

${({\beta }^{3}-{\alpha }^{3})}^{2}={(-4\sqrt{11})}^{2}=16\cdot 11=176$

So the correct answer is:

$176$

Hence, Option B is correct.

Solutions · 1