An Amazing Algebra Challenge- Give It A Try!

[@9허공]

2:45 xa^2 + (2x^2 + 1)a + x^3 - 1 = xa^2 + (2x^2 + 1)a + (x - 1)(x^2 + x + 1)
after factorization, [ a + (x - 1) ] * [ xa + (x^2 + x + 1) ] = 0
=> a = 1 - x or a = -( x^2 + x + 1)/x

[@Quest3669]

eqn can turns to
(x-1+√5)( x^2+1+x+√5x)= 0 gives X= (1-√5); {-(√5+1)+-(2+2√5)^1/2}/2 solns.

[@RealQinnMalloryu4]

➖ x^3/2^3 1x^1+2^3 x^1/2^1+1^3 (x ➖ 3x+2).

[@Fjfurufjdfjd]

Η δοθεισα γραφεται:
χ^3+2(5)^(1/2)χ^2+5χ+(5)^(1/2)-1=0
χ[χ+(5)^(1/2)]^2+(5)^(1/2)-1=0
Θετω χ+(5)^(1/2)=ψ και εχω
[ψ-(5)^(1/2)]ψ^2=1-(5)^(1/2).
ψ^3-(5)^(1/2)ψ^2+(5)^(1/2)-1=0
(ψ^3-1)-(5)^(1/2)(ψ^2-1)=0
(ψ-1)(ψ^2+ψ+1)-(5)^(1/2)(ψ+1)(ψ-1)=0
(Ψ-1)(ψ^2+ψ+1-(5)^(1/2)ψ-(5)^(1/2))=0
ψ=1 ή ψ^2+[1-(5)^(1/2)]ψ+1-(5)^(1/2)=0
χ=[-(5)^(1/2)-1+ -[2+2(5)^(1/2)]^(1/2)]/2 ή
χ=1-(5)^(1/2)

[@giuseppemalaguti435]

Sxolgo i calcoli .x^3+2√5x^2+5x+√5-1=0...(x-(1-√5))(x^2+(1+√5)x+1)=0

[@RajeshKumar-wu7ox]

X=1-√5