Question:

If1√αand1√βare the roots of the equation,ax2+bx+1=0(a≠0,a,b∈R), then the equation,x(x+b3)+(a3𕒷abx)=0has rootsα3/2andβ3/2αβ1/2andα1/2β√αβandαβα𕒷/2andβ𕒷/2If1√αand1√βare the roots of the equation,ax2+bx+1=0(a≠0,a,b∈R), then the equation,x(x+b3)+(a3𕒷abx)=0has rootsα3/2andβ3/2αβ1/2andα1/2β√αβandαβα𕒷/2andβ𕒷/21√α1√α1√α1√α1√α1√α1√α111√α√α√α√√αααα1√β1√β1√β1√β1√β1√β1√β111√β√β√β√√ββββax2+bx+1=0ax2+bx+1=0ax2+bx+1=0aax2xxx222++bbxx++11==00(a≠0,a,b∈R)(a≠0,a,b∈R)(a≠0,a,b∈R)((aa≠≠00,,aa,,bb∈∈RR))x(x+b3)+(a3𕒷abx)=0x(x+b3)+(a3𕒷abx)=0x(x+b3)+(a3𕒷abx)=0xx((xx++b3bbb333))++((a3aaa333−󔼩aabbxx))==00α3/2andβ3/2α3/2α3/2α3/2α3/2ααα3/23/23/23333////2222β3/2β3/2β3/2β3/2βββ3/23/23/23333////2222αβ1/2andα1/2βαβ1/2αβ1/2αβ1/2ααβ1/2βββ1/21/21/21111////2222α1/2βα1/2βα1/2βα1/2ααα1/21/21/21111////2222ββ√αβandαβ√αβ√αβ√αβ√αβ√αβ√√αβαβααββαβαβαβααββα𕒷/2andβ𕒷/2α𕒷/2α𕒷/2α𕒷/2α𕒷/2ααα𕒷/2𕒷/2𕒷/2−�////2222β𕒷/2β𕒷/2β𕒷/2β𕒷/2βββ𕒷/2𕒷/2𕒷/2−�////2222Aα3/2andβ3/2α3/2andβ3/2α3/2α3/2α3/2α3/2ααα3/23/23/23333////2222β3/2β3/2β3/2β3/2βββ3/23/23/23333////2222Bαβ1/2andα1/2βαβ1/2andα1/2βαβ1/2αβ1/2αβ1/2ααβ1/2βββ1/21/21/21111////2222α1/2βα1/2βα1/2βα1/2ααα1/21/21/21111////2222ββCα𕒷/2andβ𕒷/2α𕒷/2andβ𕒷/2α𕒷/2α𕒷/2α𕒷/2α𕒷/2ααα𕒷/2𕒷/2𕒷/2−�////2222β𕒷/2β𕒷/2β𕒷/2β𕒷/2βββ𕒷/2𕒷/2𕒷/2−�////2222D√αβandαβ√αβandαβ√αβ√αβ√αβ√αβ√αβ√√αβαβααββαβαβαβααββ?