Let α 1 , α 2 , … , α 7 α 1 , α 2 , … , α 7 be the roots of the…

Given:

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

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

So, $x=0$ is one of the root.

Now,

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

Put $x^2=t$, then we have

$t^3+3t^2-13t-15=0$

$\Rightarrow \left(t-3\right)\left(t^2+6t+5\right)=0$

$\Rightarrow \left(t-3\right)\left(t+1\right)\left(t+5\right)=0$

So, $t=3,t=-1,t=-5$

Now we are getting 

$x^2=3,x^2=-1,x^2=-5$

$\Rightarrow x=\pm \sqrt{3}, \pm i,\pm \sqrt{5}i$

From the given condition $\left|{\alpha }_1\right|\ge \left|{\alpha }_2\right|\ge \dots .\ge \left|{\alpha }_7\right|$

We can clearly say that $\left|{\alpha }_7\right|=0$ and

and $\left|{\alpha }_6\right|=\sqrt{5}=\left|{\alpha }_5\right|$

and $\left|{\alpha }_4\right|=\sqrt{3}=\left|{\alpha }_3\right|$ and 

$\left|{\alpha }_2\right|=1=\left|{\alpha }_1\right|$

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,

${\alpha }_1{\alpha }_2-{\alpha }_3{\alpha }_4+{\alpha }_5{\alpha }_6$

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