+125 y=x3
y2−y=2⇒{y=2y=−1}
x3=2⇒{x=(2(13)×√3×i−2(13))2x=−(2(13)×√3×i+2(13))2x=2(13)}⇒{x=−0.6299605249474366+1.0911236359724287ix=−(0.6299605249474366+1.0911236359724287i)x=1.2599210498948732}
x3=−1⇒{x=−(√3×i−1)2x=(√3×i+1)2x=−1}⇒{x=−(−12+0.866025403785i)x=12+0.866025403785ix=−1}
So the real solutions are:
+125 y=x3
y2−y=2⇒{y=2y=−1}
x3=2⇒{x=(2(13)×√3×i−2(13))2x=−(2(13)×√3×i+2(13))2x=2(13)}⇒{x=−0.6299605249474366+1.0911236359724287ix=−(0.6299605249474366+1.0911236359724287i)x=1.2599210498948732}
x3=−1⇒{x=−(√3×i−1)2x=(√3×i+1)2x=−1}⇒{x=−(−12+0.866025403785i)x=12+0.866025403785ix=−1}
So the real solutions are:
+26396 Equation X^6-X^3=2
x6−x3=2we substitute and set z=x3then we have z2−z=2z2−z−2=0z1,2=1±√1−4⋅(−2)2z1,2=1±√92z1,2=1±32z1=1+32=2z2=1−32=−1x1=3√z1=3√2=1.25992104989x1=1.25992104989x2=3√z2=3√−1=−1x2=−1
+118696 Syllogist is right :))
X6−X3=2LetY=X3Y2−Y=2$Completingthesquare$Y2−Y+14=2+14(Y−12)2=94Y−12=±32Y−12=32ORY−12=−32Y=2ORY=−1X3=2ORX3=−1X=3√2ORX=−1
These are just the real solutions :)
Let me think about the complex solutions
I think they are
x=3√2e2πi3,x=3√2e−2πi3,x=e−πi3,andx=eπi3
I hope I didn't s***w that one up :/
