x^4 +ax^3 +bx^2 +cx+d=0 let x=f(t) linear perhaps t^4 +At^3 +Bt^2 +Ct+D=0 can we have 4AB=A^3 +8C solving at most a degree three polynomial ?


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Question Number 69143 by ajfour last updated on 20/Sep/19

$x^{4} + {a x}^{3} + {b x}^{2} + c x + d = 0$ $l e t x = f (t) l i n e a r p e r h a p s$ $t^{4} + {A t}^{3} + {B t}^{2} + C t + D = 0$ $c a n w e h a v e$ $4 A B = A^{3} + 8 C s o l v i n g a t m o s t$ $a d e g r e e t h r e e p o l y n o m i a l ?$
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