Let α _ 1 , α _ 2 , , α _ 7 α _ 1 , α _ 2 , , α _ 7 be the roots of...

Given:

${x}^{7} + 3{x}^{5} - 13{x}^{3} - 15x = 0$

$⇒ x\left({x}^{6} + 3{x}^{4} - 13{x}^{2} - 15\right) = 0$

So, $x = 0$ is one of the root.

Now,

\left({x}^{6} + 3{x}^{4} - 13{x}^{2} - 15\right) = 0$</p><p>Put${x}^{2} = t$, then we have</p><p>${t}^{3} + 3{t}^{2} - 13t - 15 = 0$</p><p>$⇒ \left(t - 3\right)\left({t}^{2} + 6t + 5\right) = 0$</p><p>$⇒ \left(t - 3\right)\left(t + 1\right)\left(t + 5\right) = 0$</p><p>So,$t = 3, t = - 1, t = - 5$</p><p>Now we are getting </p><p>${x}^{2} = 3, {x}^{2} = - 1, {x}^{2} = - 5$</p><p>$⇒ x = \pm \sqrt{3}, \pm i, \pm \sqrt{5}i$</p><p>From the given condition \left|{\alpha}_{1}\right| \ge \left|{\alpha}_{2}\right| \ge \ldots . \ge \left|{\alpha}_{7}\right|$

We can clearly say that \left|{\alpha}_{7}\right| = 0$and</p><p>and \left|{\alpha}_{6}\right| = \sqrt{5} = \left|{\alpha}_{5}\right|$

and \left|{\alpha}_{4}\right| = \sqrt{3} = \left|{\alpha}_{3}\right|$and </p><p>\left|{\alpha}_{2}\right| = 1 = \left|{\alpha}_{1}\right|$

So we can have,

${\alpha}_{1} = \sqrt{5}i, {\alpha}_{2} = - \sqrt{5}i, {\alpha}_{3} = \sqrt{3}$

${\alpha}_{4} = - \sqrt{3}, {\alpha}_{5} = i, {\alpha}_{6} = - i$

${\alpha}_{7} = 0$

Hence,

${\alpha}_{1}{\alpha}_{2} - {\alpha}_{3}{\alpha}_{4} + {\alpha}_{5}{\alpha}_{6}$

$= 5 - \left(- 3\right) + 1 = 9$