I was quiet sure the solution would have this cosine-ish substitution you used before in another video but couldn't remember it... I'll have to write it down for the next time...
Well finding the value of cos(π/3) =4 cos^3(π/9) -3 cos(π/9) let cos π/9 =C then 4c^3-3c=1/2 so c^3-3/4c=1/8 . Doing substitution again a^3+b^3=1/8 and ab=1/4 So then a^6-1/8a^3+(1/4)^3=0 a^3 1,2= (1/8+-√(1/64-4×1/64))/2= (1+-√3i)/16 So C=a+b= crt((1+√3i)/16)+crt((1-3i)/16) which doesn't seem particularly nice but is a real number Maybe we can artifice the relation in the same way you would identify √(A+√B)=√C+√D to reduce the radicals and have the complex part simplify
All of it is very nice and useless at the same time, unless someone wants to play some nice math tricks and substitutions. Cubic equations are solvable from about 16th century, and especially in their canonical form. This equation is nothing special and moreover it is in its canonical form, and requires five easy calculations to arrive at three real roots of it.
I was quiet sure the solution would have this cosine-ish substitution you used before in another video but couldn't remember it... I'll have to write it down for the next time...
The radical solutions to this equation involve irreducible cube roots of complex numbers.
Well finding the value of cos(π/3) =4 cos^3(π/9) -3 cos(π/9) let cos π/9 =C then
4c^3-3c=1/2 so c^3-3/4c=1/8 . Doing substitution again
a^3+b^3=1/8 and
ab=1/4
So then a^6-1/8a^3+(1/4)^3=0
a^3 1,2= (1/8+-√(1/64-4×1/64))/2= (1+-√3i)/16
So C=a+b= crt((1+√3i)/16)+crt((1-3i)/16) which doesn't seem particularly nice but is a real number
Maybe we can artifice the relation in the same way you would identify √(A+√B)=√C+√D to reduce the radicals and have the complex part simplify
I have always loved the fun in the videos: "replace a3 with c and then you will see"
Thank you for your video. How do you prove that u is in [-1,1] in the second method?
You can using complex analysis method sir.
All of it is very nice and useless at the same time, unless someone wants to play some nice math tricks and substitutions. Cubic equations are solvable from about 16th century, and especially in their canonical form. This equation is nothing special and moreover it is in its canonical form, and requires five easy calculations to arrive at three real roots of it.