What Happens When You Solve This Math Problem?

[@Quest3669]

X=(√29-1)/ 2

[@anestismoutafidis4575]

x^3+11=(464)^1/2 x^3=(464)^1/2-11 x^3=(400•1,16)^1/2 -11 x^3=20•(1,16)^1/2 -11 x^3=20•1,077 -11 x^3=21,54 -11 x^3=10,54 x=(10,54)^1/3 x=2,192

[@RashmiRay-c1y]

We have x^3+11 = 4√29. Let x = a√29 + b > 29a^3+3ab^2=4, 87a^3b + b^3=-11 > a=1/2, b= -1/2 > x=1/2(√2-1).

[@sarantis40kalaitzis48]

Excellent, but you must write instead of sqrt2 correct sqrt29.

[@davidseed2939]

so a(29aa+3bb)=4 b(87aa+bb)=-11 but how do you arrive at aa=bb 32aaa=4, 8aaa=1, a=1/2 88bbb=-11, bbb=-1/8, b=-1/2

[@RashmiRay-c1y]

@@davidseed2939 Through inspired guesswork and tedious trial and error. I am sure there is a more systematic way, but I just don't know it.

[@RealQinnMalloryu4]

{x^3+x^3 ➖ }+{121+121 ➖ }={x^6+242}=242x^6 1^4^1x^6 4x^2^3 2^2x^1^3 1^2x1^3 2x^3 (x ➖ 3x+2).

[@Fjfurufjdfjd]

Εστω χ=[-11+4(29)^(1/2)](1/3)>0 και ψ=[-11-4(29)^(1/2)]^(1/3)

[@rabotaakk-nw9nm]

1:05 x³=4vʼ29-11= =(32vʼ29-88)/8= =(29vʼ29+3vʼ29-87-1)/8= =(vʼ(29³)-3•29+3vʼ29-1)/8= =((vʼ29)³-3(vʼ29)²•1+3vʼ29•1²-1³)/8= =(vʼ29-1)³/2³=((vʼ29-1)/2)³ => => x=(vʼ29-1)/2 😁