Using Vieta's formulas In a Cubic Equation

preview_player
Показать описание
🤩 Hello everyone, I'm very excited to bring you a new channel (SyberMath Shorts).
Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡

If you need to post a picture of your solution or idea:

#Algebra #polynomialequations #polynomials #vietasformulas
via @YouTube @Apple @Desmos @NotabilityApp @googledocs @canva

PLAYLISTS 🎵 :
Рекомендации по теме
Комментарии
Автор

erm... watched it till 2:08, then thought up the following:
Use the 1st Formula only!
since m+n+k=0 it follows
m+n=-k, m+k=-n, n+k=-m.
the expression we have to find is just (-m)^3 + (-n)^3 + (-k)^3
since m, n, k are solutions to the given cubic, for all x in { m, n, k }
x^3 + 2021x + 1 = 0, i.e. x^3 = -2021x - 1
the expression we have to find is thus equal to
-( m^3 + n^3 +k^3 ) = 2021*(m+n+k) +3 = 3
done!


After posting this, I will watch the 12 Minutes left!

(edited for grammar and a missing "-"-sign!)

karl
Автор

Third method here.
m+n+k=0
m^3+n^3+k^3 = (m+n+k)(………………..) +3mnk= 0*( ……………) =3mnk=-3
Substituting sum of terms in terms the remaining term

=-(sum of cubes)=-(-3)=3

ManjulaMathew-wbzn
Автор

Your second method could be much easier from the step:
Required expression =-(sum of cubes of toots ) =-(sum of roots)(…….) -3(product of roots)
=-(0)*(………) -3*(-1)=3

ManjulaMathew-wbzn
Автор

One thing to note: The discriminate of this depressed cubic is (p/3)^3 + (q/2)^2 where p=2021 q=1 is positive. So there is
1 real root and 2 which are complex conjugates. Let m be the real root and n=a+ib k=a-ib. Then based on the Vieta formulas
a=-m/2 and m*((m^2)/4 + b^2) = -1 So you can solve for b in terms of m. With all the roots expressed in terms of m
this may simplify the calculations. Just thought this might be another approach.

allanmarder