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Question Number 104037 by bramlex last updated on 19/Jul/20

(1) { ((x^3 +y^6  = 91)),((x+y^2  = 7 )) :}  find x−y^6  .  (2) 2a+(2/a) = 8 ⇒ ((a^6 +1)/a^3 ) ?

(1){x3+y6=91x+y2=7findxy6.(2)2a+2a=8a6+1a3?

Answered by bemath last updated on 19/Jul/20

(2) a +(1/a)= 4 ⇒((a^2 +1)/a) = 4  a^2 +1 = 4a ⇒(a^2 +1)^3 =64a^3   ⇒a^6 +3a^4 +3a^2 +1 = 64a^3   a^6 +1 = 64a^3 −3a^4 −3a^2   then ((a^6 +1)/a^3 ) = ((64a^3 −3a^2 (a^2 +1))/a^3 )  = ((64a−3(a^2 +1))/a) = 64−3(((4a)/a))  = 64−12 = 52 ■

(2)a+1a=4a2+1a=4a2+1=4a(a2+1)3=64a3a6+3a4+3a2+1=64a3a6+1=64a33a43a2thena6+1a3=64a33a2(a2+1)a3=64a3(a2+1)a=643(4aa)=6412=52

Answered by bramlex last updated on 19/Jul/20

(1) (x+y^2 )^3  = 7^3  ⇒x^3 +3x^2 y^2 +3xy^4 +y^6   = 343   ⇒91+3x^2 y^2 +3xy^4  = 343  3xy^2 (x+y^2 ) = 252 ⇒3xy^2 =((252)/7)  xy^2  = 12 ⇒y^2 = ((12)/x)  substitute to x+y^2 =7  ⇒x+((12)/x) = 7 ⇒x^2 −7x+12=0   { ((x=3 ∧y^2 =4)),((x=4∧y^2 =3)) :}  then x−y^6  = 3−4^3  or 4−3^3

(1)(x+y2)3=73x3+3x2y2+3xy4+y6=34391+3x2y2+3xy4=3433xy2(x+y2)=2523xy2=2527xy2=12y2=12xsubstitutetox+y2=7x+12x=7x27x+12=0{x=3y2=4x=4y2=3thenxy6=343or433

Answered by Dwaipayan Shikari last updated on 19/Jul/20

2(a+(1/a))=8  (a+(1/a))^3 −3(a+(1/a))=a^3 +(1/a^3 )  64−12=((a^6 +1)/a^3 )⇒52

2(a+1a)=8(a+1a)33(a+1a)=a3+1a36412=a6+1a352

Answered by OlafThorendsen last updated on 19/Jul/20

1) x = 7−y^2   (7−y^2 )^3 +y^6  = 91  7^3 −3.7^2 y^2 +3.7y^4 −y^6 +y^6  = 91  3y^4 −21y^2 +49 = 13  3y^4 −21y^2 +36 = 0  Δ = (−21)^2 −4×3×36 = 9  y^2  = ((−(−21)±3)/(2×3)) = 3 or 4  a) y^2  = 3  x = 7−3 = 4 and y^6  = 27   x−y^6  = 4−27 = −23  b) y^2  = 4  x = 7−4 = 3 and y^6  = 64  x−y^6  = 3−64 = −61  S = {−61;−23}

1)x=7y2(7y2)3+y6=91733.72y2+3.7y4y6+y6=913y421y2+49=133y421y2+36=0Δ=(21)24×3×36=9y2=(21)±32×3=3or4a)y2=3x=73=4andy6=27xy6=427=23b)y2=4x=74=3andy6=64xy6=364=61S={61;23}

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