Good❤

Good

Polynomials with order 3 have at most 3 solutions. Arranging the eqn in another order: x^3 - 27 = 0 x^3 - 3^3 = 0 (x - 3) (x^2 + 3x +9) = 0 Apparently, x = 3 and there are 2 more complex solutions which can be obtained by quadratic formula. DONE😊

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Perfect math channel, but please put the dividing line (supra, over) next to the equal sign, that's how it's done :)

just 3 seconds 😊

Use Euler's formula for questions like this. (x/3)^3 = 1 = e^(2πi)=> roots e^0, e^(2πi/3), e^(4π/3). Solutions 3 x the above list so 3, 3(-1 +√3 i)/2, 3(-1 - √3i)/2.

In my simple mind, the answeres would be 3, and absolute value of 3 and -3, the remainder of the solutions seem to be changing the equation. Someone who is smarter than I am, please explain why you would care about creating a negative... this seems overly complicated for the sake of complicating it.

Only one cube root is 3...

X=3

Simply don't waste your time & viewer's time by writing many steps unnecessarily.

All solutions are 3, 3*e^(2/3pi* i) and 3 *(e^( 4/3pi* i)

Help (x+y)^x^1/2=(x-y)^y^1/2

2 х 2 = Х. Х = ?

Read the question. How many roots. The answer is 3. And you want to waste our time. Go attend English classes

X=3