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[ritwikgupta]

0 was obvious but proving it was tricky. Setting up a sum of three squares was smart.
Zamir, Imaginary exponent of a positive integer does not give a negative I think?

[ernestdecsi]

Nice solution. I wouldn't have thought that multiplying by 2 could work such "miracles" :-)

[renesperb]

Here the only question is : is the obvious solution x = 0 the only one ?

[НиколайНиколай-жт]

Решение некорректное. Даже если мы ищем решение в действительных числах. Автор ролика показал, что никакие действительные a и b кроме 1 и 1 не дают решения уравнения, где сумма 3х полных квадратов. При этом он отбрасывает все комплексные значения a и b.
Но a и b это не окончательное, а промежуточный результат. Нас интересует их произведение. Т.к. 6^x = a*b.
Но из комплексных a и b может получится действительное произведение ab (если они комплексно сопряженные).

[vinayseth]

05:33 Why are these "perfect squares" though?

[QuentinStephens]

x = 0 is a trivially obvious solution. But you are not demonstrating that there are no complex solutions.

[corvanjer]

Looks great, one critique for the videos, is it ok to show the initial equation at the end with the solution, and maybe with the substitution for x.

[БелАлекс]

Ну вообще 0 был очевиден вначале. Когда 3 слагаемых с плюсом, 2 слагаемых с минусом могут давть единицу? Но вот чтобы доказать, что это решение единственное надо действовать как в ролике.

[НиколайНиколай-жт]

The solution is incorrect. Even if we are looking for a solution in real numbers. The author of the video showed that no real a and b except 1 and 1 give a solution to the equation, where the sum is 3 complete squares. In doing so, he discards all complex values of a and b.
But a and b are not the final result, but an intermediate result. We are interested in their product. Because 6^x = a*b.
But from complex a and b a real product ab can be obtained (if they are complex conjugate).

[anatolykatyshev]

Really nice! Make two variables instead one and solve - is non standard idea!

[PJR]

It's wrong, you cannot say (a-b)^2 >=0 (also it's not possible assume that for the others) you are missing the complex space with this assumtion and it only works in this particular case, propper way is using logaritms

[jc]

Log everything and x*log(36)/log36=0 so x=0

[krishnavrsg]

If you know the simple fact that any number raised to the power of zero is one, this becomes simple.

[lolserm]

Dude you could have solve that in just one line. The obvious answer from the beginning was x=0

[bishtkunal]

This question has an oral solution, and u r right it's 0

[matiasgualco]

0 si x=0 todos los terminos dan 1 y habria 3 positivos y 2 negativos 3-2=1

[drmjruff]

This is a test of innate ability. Recognizing the "trick" was the test. X=0. No paper or pencil.

[JimKirksHandsomeBrother]

That’s a no pencil problem.
It’s obviously zero.

[petrushkamagic_]

можно было записать всё с виде 2^x черёд логорифмы и тупую решить

[ГалиаскарГибадуллин]

Regular exercise in Russian school program