x^3 -27 = (x-3)(x^2 +3x +9) first root : x=3 consider x^2+3x +9 =0 a=1 b=3 c=9 x=( -3 +/- (9-36)^(1/2))/2 second root x = -3/2 - (3^(3/2) / 2 )i third root x = -3/2 +(3^(3/2) / 2) i Thank you for this work space
@Grim_Reaper_from_Hell4 ай бұрын
Once you get 3 the other 2 solutions are simply 3*e^(±2πi/3). (e^(2π/3))³=e^(2π)
@kpdywo8484 ай бұрын
Easy: x^3=27 in complex form : 27 exp[2k*i Pi[. So, 3 roots are: 3 exp[2k*i Pi/3]: for k=0 Root = 3, for k=1 root = -3/2+i Sqrt(3)/2, for k=2 root =-3/2-i Sqrt(3)/2
@marianondrejkovic20844 ай бұрын
Imaginary parts multiplied by 3 forgotten
@piotrstrzelczyk50134 ай бұрын
@@marianondrejkovic2084 :D
@motivationer7184Ай бұрын
Simply saying 3,3omega, 3 omega square.
@tassiedevil22004 ай бұрын
This is one case where the aficionados of the polar approach could quickly get to the same answer since most would know Cos[2 pi/3] = -1/2, Sin[2 pi/3] Sqrt[3]/2 etc.
@joeldurugbo31474 ай бұрын
So simply multiply the real root by the primitve cube roots of unity to get the two complex roots
@donaldasayers4 ай бұрын
Olympiad problem? I think not, O-level further maths for me. Roots of a number are always distributed equally around a circle in the Argand plane.
@Grim_Reaper_from_Hell4 ай бұрын
The word Olympiad is a clickbait
@SPDLand4 ай бұрын
Huh?
@Roger_Gadd11 күн бұрын
Yep. Much quicker. Let x=r.cis(theta). We know r=3, and the three values of theta are obvious.
@MadScientyst4 ай бұрын
To the uninitiated, once the word ALL roots are mentioned in a Math problem, the alarm bell sould sound off that the soultion set MUST involve the Complex Domain, eg. 1 can also be represented as 1+0i & even 0 as 0+0i, so a typical algebraic equation can have both Real & Complex solutions....🤔
@YahyaMohammed-y4v16 күн бұрын
According to rules of the exponential function: x = 27 exponent 1/3
@vfbfvgdgb-zr5kq4 күн бұрын
I think that x=3😅
@DedMatveev4 ай бұрын
The author, as usual, "forgot" to say when posing the problem in which SET he wants to search for x: among real or complex values. Both the solution method and the answer depend on this.
@joso55544 ай бұрын
Thanks for pointing this. These YT videos almost never bother to mention this essential input.
@stvp684 ай бұрын
“Find all roots” implies complex roots, doesn’t it?
@bendunselman4 ай бұрын
@@stvp68No, not necessarily. There may well be solutions in other algebras over other rings/fields/domains. For instance in a Clifford algebra over the quaternions, octonions, split complex numbers, extended complex numbers, dual numbers or a Z[27] field. (But I did not think of solving any of those) The maths world has a lot more space to navigate....
@stvp684 ай бұрын
@@bendunselman ah, okay, thanks!
@kitkann78654 ай бұрын
@@joso5554លៀ😢៩ផ៥ហឱអ:យ🎉លេីៀ+ឱ🎉ន🎉ៀល៤ឬៀនរ៥ល😊្
@PravinKumar-v6x3 ай бұрын
Is there anyone from INDIA❤
@ClaudioBrogliato4 ай бұрын
If I remember well putting those solutions on a complex plane should result in the vertex of an equilateral triangle.
@ClarkPotter4 ай бұрын
That's pleasing.
@parthasarathidas23603 ай бұрын
Using cube roots of unity concept . its in iit jee syllabus
@LloydCash-he1qv4 ай бұрын
x^3 = 27 x^3 = 3^3.e^(i.2nπ) x = 3.e^(i.2nπ/3) x = 3 or x = -3/2 + i.3√3/2 or x = -3/2 - i.3√3/2
@LloydCash-he1qv4 ай бұрын
The complex solution is best arrived at using Euler's formula. The author's algebraic method is long and tedious as well as essentially unnecessary!
@abneryokum3 ай бұрын
@@LloydCash-he1qv This guy is a Bozo. All his solutions are long and tedious.
@RealQinnMalloryu44 ай бұрын
(x ➖ 3ix+3i )
@1234larry14 ай бұрын
Once you get the difference of squares, difference of cubes and sum of cubes down, its not that hard.
@bjornfeuerbacher55144 ай бұрын
Not necessary for solving this equation.
@sagarrajak38264 ай бұрын
Yes one root is 3 and the other two are complex i.e complex conjugate
@Yahya2-mz3jmАй бұрын
Why he doesn't use demoaver?
@ShivamSharma204804 ай бұрын
It's simple look X^3=27 find the value of x=? 1step x^3=27, 2step x=rootunder27, 3step x=3 because 3 is the cube number of 27
@mohamad-hu9vv3 ай бұрын
Great!
@Misha-g3b4 ай бұрын
3, -1,5+_1,5iV3.
@minewithsayo42204 ай бұрын
Quite interesting!
@rastikigrov84542 ай бұрын
but what about the root x = -3i^2/3?
@gregoryian1083 ай бұрын
wow!
@Mister_Mouse73 ай бұрын
:/ x³=27 x=3√27=>3
@poovizhisaravanan64212 ай бұрын
3^3=27
@azhanahmedali98514 ай бұрын
You forgot that there exist somethings like log .
@bjornfeuerbacher55144 ай бұрын
Why should one use log for this equations???
@azhanahmedali98514 ай бұрын
@@bjornfeuerbacher5514 it makes it much easier than ever .
@bjornfeuerbacher55144 ай бұрын
@@azhanahmedali9851 ??? Sorry, I don't see it. Please explain to me how one solves this equation using log.
@azhanahmedali98514 ай бұрын
@@bjornfeuerbacher5514 To convert the equation x³ = 27 into logarithmic form, we can use the property of logarithms that states: logₐ(b) = c ⇔ a^c = b In this case, we can rewrite the equation as: x³ = 27 logₓ(27) = 3 Or, alternatively: 3 logₓ(x) = log₇(27) Since 27 is 3³, we can also write: 3 logₓ(x) = 3 log₇(3) logₓ(x) = log₇(3) Note that the base of the logarithm can be any positive real number, but in this case, I've used base 7 (since 27 = 3³) and base x (to maintain the original variable).
@bjornfeuerbacher55144 ай бұрын
@@azhanahmedali9851 "logₓ(27) = 3" This does not help for solving the equation. "3 logₓ(x) = log₇(27)" ??? Where did you get that from? That's simply not true! "Since 27 is 3³ Err, as soon as you recognize that, you have found the (real) solution of the equation x³ = 27. So you just demonstrated yourself that log does not help in any way for solving this equation - you only have to see that 3³ = 27, and you are done! "but in this case, I've used base 7 (since 27 = 3³)" ????? There is no reason at all for using base 7 here! Did you perhaps mean base 27?
@williamgraham24683 ай бұрын
Abraham de Moivre would like a word...
@RafaAssyifa4 ай бұрын
X=3
@mariapiazza-od8ib4 ай бұрын
EINSTEIN in the thumbnail ? 😢😢 Apmk he's NOT a mathematician, nor was he a CHAMP in maths 😮😮 PLEASE check it out.. and let us know 😊😊
@dhruvtakiar3 ай бұрын
Who cares when he was truly a genius
@vaichu44u4 ай бұрын
I just did the x^3=1 version of this today 😂😂
@tombufford1364 ай бұрын
The complex roots seem unmeaningful.
@bjornfeuerbacher55144 ай бұрын
Why?
@tombufford1364 ай бұрын
@@bjornfeuerbacher5514 X^3 is not a circular or repeating function, it tends to infinity. There is one intersection of x^3 and 27 when x = 3. the cube of a negative number is negative and 27 is positive.
@bjornfeuerbacher55144 ай бұрын
@@tombufford136 Yes so all what you have written. But so what??? What does all of that have to do with your assertion that the complex roots are "unmeaningful"?
@Parsi-Aria3 ай бұрын
۳×۳×۳=۲۷ I feel like Nikolas tesla 😊
@АндрейЛюбавин-э4щ4 ай бұрын
3
@Avallach004 ай бұрын
-3i
@Peterthepainter664 ай бұрын
3x3x3=27 that's it.
@nsy4652Ай бұрын
No way
@markphc994 ай бұрын
Be careful of what? This is a straightforward problem , not really higher maths.
@antonygattini30734 ай бұрын
😄😄🤣🤣🤣🤣😂😂
@Muslem_x5i3 ай бұрын
Anyone from syria 🤎✨
@AnhNguyenNhat-rk3mm3 ай бұрын
unvaliable bommmmmmmmmmmmmmm
@halcon21344 ай бұрын
Try a limit but that has no solution
@gopagon21143 ай бұрын
X=9
@bmrw4 ай бұрын
9(3)=27 so its Simple 9+9+9=27
@szvetozargligorovics94313 ай бұрын
Hope you not teaching mathematic because your x it look like Coco Chanel sign
@zaphodbeeblebrox-fz5fh4 ай бұрын
Not an olympiad question at all. And an horrible error in line 2 on the board, because (inside the complex numbers, which are needed anyway), x = 3 is _not_ the only solution of the equation. Do not present it like this