France - Math Olympiad Question | Be Careful! 

x^6 = (x-6)^6 arg(x-6) = nπ/3+arg(x) |x|^2 = |x-6|^2 let x= a+bi |x|^2 = a^2 + b^2 = |x-6|^2 = a^2 + b^2 -12a + 36 0 = 36-12a a=3 that gives the real component, and that x-6 = -3 + bi with some basic trig knowledge, you can infer that 0 -3 = 3/2 - bsqrt(3)/2 3 = bsqrt(3) b = sqrt(3), giving x = 3 + sqrt(3)i for n = 3, that imples -3+bi = (3+bi)e^πi, which is b=-b, so trivial to show x = 3 with multiplicity 2

I’ll guess x=3 before watching this video

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Mr. Higher Math : Very nice solution presented here. The idea of factoring by difference of squares as a first step is really good. Here is one other nice trick that will shorten the work even more : The quantities (x) and (x-6) that are being raised to the 6th power are not symmetrical, but if we choose another variable "y" that lies exactly halfway between these two, then we can make the equation symmetrical. Therefore, use y = x - 3 . Then, x = y + 3 , and x - 6 = y + 3 - 6 = y - 3 . Now the equation to solve is symmetrical : (y+3)^6 = (y-3)^6. Now apply your trick of factoring by the difference of squares, and the algebra work becomes much shorter. In fact it reduces to this : [ 2y^3 + 54y ] [ 18y^2 + 54 ] = 0 . This becomes : y [ y^2 + 27 ] [ y^2 + 3 ] = 0 . We quickly get : y = 0 OR y = +/- sqrt(-27) = +/- 3sqrt(3)i OR y = +/- sqrt(-3) = +/- sqrt(3)i . Then, with x = y + 3 this gives the same 5 solutions that you obtained.

It's too long. Elevating to the power 6 is useless. In fact, since x can't be 6 we can divide by (x-6)^6. So we get the equation: (x/ (x-6)) ^6 =1. Then we get two equations in IR x/x-6 =1 or -1 and four equations in C : x/ x-6 = exp(2i* PI *k/6) where k=1,2,4,5. The first equation x/x-6=1 doesn't have any solution, because x=x-6 doesn't.

2:40 it is NOT “times another parenthesis.” It is “times the quantity” which is inside the parentheses. You NEVER multiply parentheses, only quantities.

Just to be clear, even if the equation appeared with the 6th grade, this is a 5th equation, so it has 5 solutions in total.

Задача поставлена некорректно. В начале не было сказано, что надо найти ВСЕ КОМПЛЕКСНЫЕ решения. Если по умолчанию предполагается искать ДЕЙСТВИТЕЛЬНЫЕ корни, то идея извлечь корни 6 степени из обеих частей вполне здравая. Только нужно учесть, что степень чётная и получить равенство модулей.

Starting at 5:07, shouldn't that result in - 6, not +6? I see a - left there after the two x's subtract.

Why we can't square root 6 both side.. and find ans

We can see that x can't be equal to the x-6 then x=-(x-6) And we get easy equation 2x=6 x=3

Why not elevate each side of the equation to the power 1/3. We ended up with x**2=(x-6)**2. Developping the right side, we'll have x**2=x**2-12x+36. Simplifying: 12x=36 then x=3. Simple!

what is the role of this (6) in the factorisation because éphémère didn’t even use it to find a solution (isn’t he supposed to multiply it to the parenthese…)

Wait you should be geting exactly 6 roots right but we got only 5 roots please explain

You are such a patient teacher. Congratulations

All five roots have to be 6 no?

It is NOT x minus x minus six. It is x minus the quantity x minus six.

Некоторые могут спросить почему в решении получилось 5 корней ведь в уравнение входит x в шестой степени. Дело в том, что, если разложить правую часть с помощью бинома Ньютона, то x в шестой степени в левой и правой частях уравнения сократятся, и самая высокая степень x будет пятая, и, следовательно, решение будет иметь 5 корней.

3, when x is from R.

My approach to this is to first note that x=0 is not a solution, so we can divide the equation out by 1/x^6. This gives us: (1-6/x)^6 =1 Solutions of this are of the form 1-6/x = p^n or x = 6/(1-p^n) where p^6 = 1=e^(2*π*i), or p=e^(2*π*i/6) and n=0..6 Note that n=0 is a superfluous root, corresponding to the solution of x=(x-6). So it's actually only n=1..5 that are solutions We can improve symmetry of the algebra by defining m=n-3, and noting that p^m=-1 x=6/(1-p^(m-3)) = 6/(1-p^m /p^3) = 6/(1+p^m) = 6* p^(-m/2) / [2*cos(m*π/6)] = 3*[cos(m*π/6) - i*sin(m*π/6) ] /cos(m*π/6) = 3 - 3*i*tan(m*π/6) for m=-2..2 tan(m*π/6) is 0, +-1/sqrt(3) and +-sqrt(3) for m=0, +-1 and +-2 respectively Note the switch to m is entirely optional. The same results can be found without making the substitution, but a little more baggage carries through the algebraic manipulation.

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Shouldn’t there be six roots?

X^2=(x-6)^2 X^2=X^2-12X+36 12x=36 Therefore x=3

Are we just not going to talk about how there were eleven 3s and no other numerals in the final five answers

Nice problem.Thanks!

Nice level of detail there. Not just giving up when you get one real answer like some would.

Excellent proof. Thank you.

(x ➖ 3x+2)

3 !!

Good Soliton Thanks 🙏❤

If a^m=b^m then minimum one value of a =b or a= -b from indices therefore x=x-6 which impossible so x= -(x-6) then x=3