I agree, and don’t know why the instructor did not mention, much less discuss why x cannot be 0 in the cubic equation. X looks like a solution after you factor it as you did, but when you let x be 0 in the original equation, you end up with 0/0 which is undefined.
@@monroeclewis1973 ...because it abvious it can be x³/(3x) = 6 -> x³ = 18x -> x³ - 18x = 0 -> x(x² - 18) = 0 Obviously the the solutions of x are, x = [0, -3√2, 3√2] its a continuous equation and as such has a solution for x=0
(x.x.x)/(x + x + x) = 6 → where: x ≠ 0
x³/(3x) = 6
x³ = 18x
x³ - 18x = 0
x.(x² - 18) = 0 → recall: x ≠ 0
x² - 18 = 0
x² = 18
x = ± √18
x = ± 3√2
I agree, and don’t know why the instructor did not mention, much less discuss why x cannot be 0 in the cubic equation. X looks like a solution after you factor it as you did, but when you let x be 0 in the original equation, you end up with 0/0 which is undefined.
@@monroeclewis1973 ...because it abvious it can be
x³/(3x) = 6 ->
x³ = 18x ->
x³ - 18x = 0 ->
x(x² - 18) = 0
Obviously the the solutions of x are, x = [0, -3√2, 3√2]
its a continuous equation and as such has a solution for x=0
X = 3×2^.5
This is a very simple math problem and not a Math Olympiad Problem.