Solving x(x(x+1)+1)+1=0

[@bprpmathbasics]

How to really solve sin(x^2)=sin(x)
ua-cam.com/video/gCOWo13xxzc/v-deo.html

[@amlananshumatsahu5850]

Expand it: x³ + x² + x + 1 = 0
Write it as (x⁴ - 1)/(x-1) = 0
So, x⁴ - 1 = 0, x ≠ 1
x⁴ = 1, x ≠ 1
Therefore x = -1, i, -i

[@kdog3908]

Thank you!

[@alphazero339]

You can put how many iterated x you want and I think it will always be easy to solve using geometric series
Doing that you find if theres N many x's in the equation (now we have 3) the solutions are all (N+1)th roots of unity except 1

[@ZDTF]

What's a geometric series

[@ZDTF]

Alpha zero😮

[@alphazero339]

@@ZDTF you can google it its very useful sum of the form xⁿ+x²ⁿ+x³ⁿ+x⁴ⁿ+x⁵ⁿ...
Usually we use n=1 and the key is that if you multiply the series by 1-x, first terms cancels second, second cancels third, third cancels fourth and so on..

[@rainerzufall42]

1:10 🛑‼x³ + x² + x + 1 = 0. If we had (x - 1) times this one,
it would be x^4 - 1 = 0 or x^4 = 1 with the solutions { +/-1, +/-i }.
Thus the original equations has the solutions { -1, +/-i }. ✅🏁

[@Brid727]

what i did is i distributed the x all the way at the front of the parentheses without expanding the inside and by doing so i immediately noticed that it factors to (x^2+1)(x+1) and ofc the answers are -1, +/-i

[@Mathsfighters0707]

🎉🎉🎉

[@Starstrike149]

But -1 works

[@ΓιΘωεγ]

I have a big π²

[@subavajra]

1 minute ago