Solving A Nice Cubic System in Two Ways

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This problem is from "800 лучших олимпиадных задач по математике"
#ChallengingMathProblems #PolynomialSystems

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Or you could just subtract the equations, and factor out X + Y. Then it's done. And no weird "tricks" necessary.

frentz
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3rd approach: you can also subtract the second equation from the first, obtaining x^3+y^3-3xy(x+y)=0. Then, write x^3+y^3 as (x+y)^3-3xy(x+y), and get, after trivial manipulations, (x+y)[(x+y)^2-6xy]=(x+y)(x^2+y^2-4xy)=0, from which you get y=-x and y=2+-sqrt(3), and then on like in the 1st approach...

andreaambrosio
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You can bring IMO problems, that are the most challenging problems in the world

shankhadeepghosh
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1) r=sqrt(2) not 1/sqrt(2)
2) sin(7*pi/12) not sin(7*pi/2)
3) x-yi = r( cos(a) + isin(a) ) => x=r*cos(-a) and y=r*sin(-a) not y=r*sin(a) !

ukaszsimbiga
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On 2nd approach why r=1/sqrt2 thats not sqrt2 ?

yuycgddxh
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Very interesting. Using the complex numbers to find the real solutions. I did not know that was a thing.

davinheagertans
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No need to use "k" method, just equalize two eqns, factorize x+y

Also, a cubic eqn system normally has 3 pairs of solutions. The answer given here is actually incomplete.

In addition to x=-y, so y=-x,
another 2 pairs of solutions are (pair signs opposite):
x=(2±√3)y, so y=(2-/+√3)x and
y=(2±√3)x, so x=(2-/+√3)y

xz
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If X = xA and Y = yA is solution then X = - yA and Y= xA is also solution (we obtain each expression from the other by making the substitution) ---> xA^3 -3xA*xA^2 = -2xA^3 = 1 ---> xA = -(1/2)^1/3 and yA = (1/2)^1/3 and reverse
is it rigorous ?!

WahranRai
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Solving it with the complex numbers trick was really cool, great job!😃💯👌

yoav
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Hi, I have an interesting problem I came up with while learning integrals, I tried to calculate the area between 2^x and cos(x), the biggest problem was the calculation of the interception point, I asked my math teacher about it, and she didnt know either, I would be greatful for any help, Wolfram Alpha gave me the answer but not the explanation

tomasgalambos
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In the second method, I think exists a bug.
When is said that
1-i=1/sqrt(2)*(cos...
Should say
1-i=sqrt(2)*(cos...
No?

rafaeldominguez-castro
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-- good job on the handwriting this time

frentz
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Amazing video Syber, however I think the part with ( x-yi) = …….. after using De Moivres Theorem the imaginary part your supposed to set it equal to NEGATIVE y.

moeberry
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Oh no, I just had a control on the trigonometric form of complex nomber, this video was help me 😭

damiennortier
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Neyi soruyor. X mi 3x mi
X küp mu. Soruyu sormadan direkt çözüme niye geçiyorsun

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