I did it with basics only: ∛(∛2-1) = 1/∛(∛4+∛2+1) - cubic root conjugate. We want to get rid of the outer root, so our only hope is if the trinomial is a "perfect cube" of something - which it is: (∛2+1)³ = 3∛4+3∛2+3, so the very first shot is lucky and we get: 1/∛(∛4+∛2+1) = ∛3/(∛2+1). This implies: (∛2+1)(∛a+∛b+∛c) = ∛3. Now let's set a = 3x, b = 3y, c = 3z. Then: (∛2+1)(∛x+∛y+∛z) = 1 which is "almost" the sum of cubes expansion. If we let x = 4, y = -2, z = 1 we get (∛2+1)(∛x+∛y+∛z) = (∛2)³+1³ = 3, so dividing both sides by 3 we find x = 4/27, y = -2/27 and z = 1/27 so finally a = 4/9, b = -2/9, c = 1/9
1. Good handwriting. 2. Great energy! 3. Great sound quality, and your voice is very effective for audio/video. 4. Well organized. It is deceptively hard to do math this complex on a video but you handled it like a champ! Good job, sir
This RHS is out of proof, being ab initio a real source of confusion. It is only a final of the LHS.So,...work must be done only in LHS. Anyway...the proof is very interesting. Newton is a smart teacher who explains with the necessary patience. Congratulation prof. Prime Newton.
Well... Wait a moment please, sir... :D We didn't have ANY constraints EXCEPT that all three values should be element of rationals... Am i right or is this justified? I have a drastically easier approach to that problem, you know... Since we have three variables, but only one defining equation... you know... ...we are entitled to chose arbitrary values for two of the variables, as long as the remaining part(s) of the expression keep the ability to express the desired result. Since all three expression parts on the RHS are equivalent in their value range and all provide equally for a sufficient range, ... I use my freedom of choice to set a = 0, b = 0, c = 2^(1/3) - 1. And am done. No rule broken. Task giver satisfied - i hope :D
@@PrimeNewtons The six different solutions are just that a, b, and c are interchangeable. The answer should have been given that (a,b,c) belongs to the set {1/9, -2/9, 4/9}.
Remember, x is not a variable here. It is a specific constant. So the equation is only true here because of that substitution made. Your name is glorious ✨️
Sir ,I am a math lover.and running78yrs using bifocal spects. I hope u made this video for those who watch it in their mmobiles. As I was a teacher by profession I appreciate ur attitude as if u teaching in a classroom. But it's very difficult to follow ur board work as it's not even visible in mobiles ,I sa,y. Please try to write a little bold . Thank u.😂 Pl
