Another way: Euler's formula gives the solution. In polar form it's six vectors each with magnitude 4 and angles 0, 60, 120, 180, 240, and 300 degrees 🙂
Multiply both sides by x, you get 64 is equal to x squared over 4 multiplied by x squared over 4 multiplied by x squared over 4. Multiply by 4 on both sides to cancel out the 4's. You get 64 times 4 which is 256 is equal to x raised to the 6 when you Multiply them together, then take the 6th root of the both sides, you get x is equal to 6th root of 256. X is equal to 2.51984.
Guys: you gotta remember the fundamental theory of algebra, and complex numbers. Yall always forget that and brag about finding the answer "so quickly" ehen you havent found the answer at all
Ok the solution makes more sense, I didn't even think about making it all powers.
Another way: Euler's formula gives the solution. In polar form it's six vectors each with magnitude 4 and angles 0, 60, 120, 180, 240, and 300 degrees 🙂
Multiply both sides by x, you get 64 is equal to x squared over 4 multiplied by x squared over 4 multiplied by x squared over 4. Multiply by 4 on both sides to cancel out the 4's. You get 64 times 4 which is 256 is equal to x raised to the 6 when you Multiply them together, then take the 6th root of the both sides, you get x is equal to 6th root of 256. X is equal to 2.51984.
Just to clarify at the beginning the x's canceled out and I multiplied 4 times 4 times 4 to get 64 if your confused.
±4, -2±2iV3, 2±2iV3.
Молодец, Миша, совершенно верное решение
@СергейИванчиков-р8д Спасибо, Серёжа!
Guys: you gotta remember the fundamental theory of algebra, and complex numbers. Yall always forget that and brag about finding the answer "so quickly" ehen you havent found the answer at all
+ or - 4
I solve it in .5 second 😅
You didn't because you need to find *all* of the roots, not just the easiest one
4
J’ai trouvé rapidement la solution
x=+4 ou x=-4
You're missing four of the roots
x=4e^(kπi/3) ∀k∈ℤ.
You joking ? X=4 and x= -4 no need to calculation
@@Radioayandeh There are many other roots.
@Misha-g3b yes but we need only one answer
@@Radioayandeh We have to try to find all roots, if it is not difficult.