Given equation can be rearranged as x(x6+3x4−13x2−15)=0 clearly x=0 is one of the root and other part can be observed by replacing x2=t from which we have t3+3t2−13t−15=0 ⇒(t−3)(t2+6t+5)=0 So, t=3,t=−1,t=−5 Now we are getting x2=3,x2=−1,x2=−5 ⇒x=±√3,x=±i,x=±√5i From the given condition |α1|≥|α2|≥....≥|α7| We can clearly say that |α7|=0 and and|α6|=√5=|α5| and |α4|=√3=|α3| and |α2|=1=|α1| So we can have, α1=√5i,α2=−√5i,α3=√3i, α4=−√3,α5=i,α6=−i Hence α1α2−α3α4+α5α6 =1−(−3)+5=9