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Question Number 71960 by TawaTawa last updated on 22/Oct/19

Answered by mind is power last updated on 22/Oct/19

Σ((1+a)/(1−a))=Σ(−1+(2/(1−a)))=Σ−1+Σ(2/(1−a)) ((p′(x))/(p(x)))=Σ_(a∈Root(p)) (1/(x−a)) ⇒Σ_(i=1) ^3 (1/(1−x_i ))=((p′(1))/(p(1)))=((3−1)/(−1))=−2 ⇒Σ_ ((1+a)/(1−a))=−2.2+Σ−1=−4−3=−7 2nd Solution just ((1+a)/(1−a))+((1+b)/(1−b))+((1+c)/(1−c))=(((1+a)(1−b)(1−c)+(1+b)(1−a)(1−c)+(1+c)(1−a)(1−b))/((1−b)(1−c)(1−a))) x^3 −x−1=(x−a)(x−b)(x−c)⇒(1−a)(1−b)(1−c)=1−1−1=−1 (1+a)(1−b)(1−c)=1+(a−b−c)−ab−ac+bc+abc (1+b)(1−a)(1−c)=1+(b−a−c)+ac−ab−bc+abc (1+c)(1−a)(1−b)=1+(c−a−b)+ab−ac−bc+abc (((1+a)(1−b)(1−c)+(1+b)(1−a)(1−c)+(1+c)(1−a)(1−b))/((1−b)(1−c)(1−a))) =((3−(a+b+c)−ab−ac−ac+3abc)/(−1)) a+b+c=0 ab+ac+cb=−1 abc=1 =((3−(a+b+c)−ab−ac−ac+3abc)/(−1))=((3−0+1+3)/(−1))=−7

Σ1+a1−a=Σ(−1+21−a)=Σ−1+Σ21−ap′(x)p(x)=∑a∈Root(p)1x−a⇒∑3i=111−xi=p′(1)p(1)=3−1−1=−2⇒∑1+a1−a=−2.2+Σ−1=−4−3=−72ndSolutionjust1+a1−a+1+b1−b+1+c1−c=(1+a)(1−b)(1−c)+(1+b)(1−a)(1−c)+(1+c)(1−a)(1−b)(1−b)(1−c)(1−a)x3−x−1=(x−a)(x−b)(x−c)⇒(1−a)(1−b)(1−c)=1−1−1=−1(1+a)(1−b)(1−c)=1+(a−b−c)−ab−ac+bc+abc(1+b)(1−a)(1−c)=1+(b−a−c)+ac−ab−bc+abc(1+c)(1−a)(1−b)=1+(c−a−b)+ab−ac−bc+abc(1+a)(1−b)(1−c)+(1+b)(1−a)(1−c)+(1+c)(1−a)(1−b)(1−b)(1−c)(1−a)=3−(a+b+c)−ab−ac−ac+3abc−1a+b+c=0ab+ac+cb=−1abc=1=3−(a+b+c)−ab−ac−ac+3abc−1=3−0+1+3−1=−7

Commented by TawaTawa last updated on 22/Oct/19

God bless you sir, thanks for your time

Godblessyousir,thanksforyourtime

Commented by TawaTawa last updated on 22/Oct/19

Sir is the answer − (7/3) ???. That is what they got here.

Siristheanswer−73???.Thatiswhattheygothere.

Commented by mind is power last updated on 22/Oct/19

no −7

no−7

Commented by TawaTawa last updated on 22/Oct/19

Ok, God bless you sir. I appreciate your time

Ok,Godblessyousir.Iappreciateyourtime

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