Nice video! As I was only able to come up with 2 methods to solve this problem, I was looking forward to learn a third method. My disappointment in this regard was well compensated by seeing the b=ka substitution in the equations with homogeneous LHS, leading to a simple cubic (I found the cube root by an inspired guess at the answer and checking, a less robust method). I do think though that in method 3 it would have been prudent to check that the result found is in fact a cube root of 2+√5.
I agree. Kinda. It's true that when dealing with complex numbers, raising both sides to any power (here 3) might introduce new solutions. However, because we eventually found a real solution, we can say with certainty that it will in fact be the correct answer
P.S. I once wrote a very similar problem for a maths competition: ³√(10+6√3)+³√(10-6√3)=? Answer is 2, as can be seen by deriving the related cubic, x³+6x-20=0 Or by finding cube root of expressions to be 1±√3
@@skylardeslypere9909 Sure, but my point is we started with an assumption about the form of the cube root, so in the end it would be prudent to verify that our assumption was correct.
Comment regarding your 3rd method: you know that a,b nust be multiples of 1/2 which differ by an integer. This is because cbrt(2+sqrt 5) is satisfied by a monic integer polynomial. Thus, it is an algebraic integer. However, the number field Q(sqrt5) has an integral basis {1,(1+ sqrt 5)/2}.
My method: this is obviously a cubic equation's solution. So go back to the cubic equation by cubing ?. We get ?³ + 3? = 4. The only real solution is ? = 1. It's quite ironic that to solve a cubic equation, we need to take the solution, return to the cubic equation, and resolve it. Or possibly go to a different cubic equation.
How do you know that t^3+3t-4=0 gives you 1 as a root? (The positive discriminant of the reduced cubic equation shows us that there is only one real root, but it does not tell us how to calculate it, that is, how to transform the cubic roots into a number without cubic radicals).
@@matheusalmeidadamata la trovo a mente si vede subito che una radice è 1,altrimenti devo utilizzare la formula risolutiva delle equazioni di 3 grado, molto complicata.... Ma fortunatamente si vede a occhio t=1,poi si abbassa il grado con Ruffini....in effetti, adesso che controllo meglio, la formula risolutiva è esattamente il calcolo di partenza.... Beh allora, il calcolo lho fatto a mente.... Spero di aver capito e di essermi fatto capire
здравствуйте Я учился в Украине.Вы очень сложно решаете.Приравняем иксу умножаем обе стороны на два внеснм двойку под корень.шестьнадцать плюсь восемь корень пять.это польный куб.
Got it using the first method!
Nice!
直接利用三次方和的二項式展開就可以了，不用那麼麻煩
you always leave the beauty and the greatness to the last method in the video
Thanks!
Nice video! As I was only able to come up with 2 methods to solve this problem, I was looking forward to learn a third method. My disappointment in this regard was well compensated by seeing the b=ka substitution in the equations with homogeneous LHS, leading to a simple cubic (I found the cube root by an inspired guess at the answer and checking, a less robust method). I do think though that in method 3 it would have been prudent to check that the result found is in fact a cube root of 2+√5.
I agree. Kinda. It's true that when dealing with complex numbers, raising both sides to any power (here 3) might introduce new solutions. However, because we eventually found a real solution, we can say with certainty that it will in fact be the correct answer
P.S. I once wrote a very similar problem for a maths competition:
³√(10+6√3)+³√(10-6√3)=?
Answer is 2, as can be seen by deriving the related cubic, x³+6x-20=0
Or by finding cube root of expressions to be 1±√3
@@skylardeslypere9909 Sure, but my point is we started with an assumption about the form of the cube root, so in the end it would be prudent to verify that our assumption was correct.
@@MichaelRothwell1 should one of the 6sqrt3 be negative?
@@skylardeslypere9909 Fixed, thanks for spotting this!
Amazing 🤩
What is name of the board that you use??
Thank you!
It's notability
Comment regarding your 3rd method: you know that a,b nust be multiples of 1/2 which differ by an integer. This is because cbrt(2+sqrt 5) is satisfied by a monic integer polynomial. Thus, it is an algebraic integer. However, the number field Q(sqrt5) has an integral basis {1,(1+ sqrt 5)/2}.
That's cool!
cbrt(2 ± sqrt(5)) = 1/2 ± sqrt(5)/2 so the answer is 1
I just rearranged the expressions in the cube roots to get cbrt(phi^3) + cbrt((-1/phi)^3)
Great Video, Method 3 is great.
method 3 might be useful in some cases, but here its just a huge overkill
Thank you! 😊
Bravo
Great problem , all method are awesome good job
Thanks a lot
My comments on the earlier video ua-cam.com/video/_eLQB5kDGhE/v-deo.html for root 5 plus 2 still apply. Good stuff.
nice video sir thanks
Most welcome
2 + 5^1/2 +2 - 5^1/2 = 4
This is actually how Cardano's formula expression looks like :)
My method: this is obviously a cubic equation's solution. So go back to the cubic equation by cubing ?. We get ?³ + 3? = 4. The only real solution is ? = 1.
It's quite ironic that to solve a cubic equation, we need to take the solution, return to the cubic equation, and resolve it. Or possibly go to a different cubic equation.
Pongo la somma =t, elevo al cubo, e lavorando un po', ha una cubica come equazione finale. t^3+3t-4=0 che mi dà come unica soluzione reale t=1
How do you know that t^3+3t-4=0 gives you 1 as a root? (The positive discriminant of the reduced cubic equation shows us that there is only one real root, but it does not tell us how to calculate it, that is, how to transform the cubic roots into a number without cubic radicals).
@@matheusalmeidadamata la trovo a mente si vede subito che una radice è 1,altrimenti devo utilizzare la formula risolutiva delle equazioni di 3 grado, molto complicata.... Ma fortunatamente si vede a occhio t=1,poi si abbassa il grado con Ruffini....in effetti, adesso che controllo meglio, la formula risolutiva è esattamente il calcolo di partenza.... Beh allora, il calcolo lho fatto a mente.... Spero di aver capito e di essermi fatto capire
@@giuseppemalaguti435 Grazie signore, buona notte.
здравствуйте Я учился в Украине.Вы очень сложно решаете.Приравняем иксу умножаем обе стороны на два внеснм двойку под корень.шестьнадцать плюсь восемь корень пять.это польный куб.
Привет. Спасибо за ответ!
cube root of negative 1 has three solutions, like every root of n-th degree has n solutions :)
No. The usual definition of cube root is that it takes a real number and returns a real number. The cube root is NOT the same as the 1/3 power.
👍
i love 3 methods videos
Definitely 1
Let cubrt(2 + sqrt(5)) + cubrt(2 - sqrt(5)) =t
Cube both side
Cube(t) = 2+sqrt(5) + 2 -sqrt(5) + 3(cubrt(4-5) x (cubrt(2 + sqrt(5)) + cubrt(2 - sqrt(5)))
Cube(t) = 2 + 2 + 3(-1)(t)
Cube(t) = 4-3t
Cube(t) +3t -4 =0
(t-1)(square (t) +t+4) = 0
t=1
Quadratic has complex soln
How did you do this factorization (from t^3+3t-4=0 to {t-1}{sqrt(t)+t+4})?
first equation... f(x)=0 = sad face ....I did not like it
Today's problem is good
How about right now?
Each cube root has three solutions once you include complex numbers, so there are actually nine solutions, not one.
Answer is 1.
This video is number 1! (And 1 and 1😃)
😍
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