Solving a Nice Cubic System in Two Ways

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[Chrisoikmath_]

My method (it is like your second one but I think it is more simple):
Step1: Adding the two equations, we conclude that x+y=3 (:1)
Step2 : Sumtracting the two equations, we conclude that x-y=1 (:2)
Step3 : Adding equations (:1) and (:2), we conclude that x=2, so y=1
Therefore: (x, y)=(2, 1)

[titassamanta]

Add them and u will get (x+y)^3 = 27. So, x+y=3. Subtracting the second equation from the first one we get x^3-y^3+3xy^2-3yx^2=1 So (x-y)(x^2+xy+y^2)-3xy(x-y)=1 => (x-y)(x^2+xy+y^2-3xy)=1 => (x-y)(x^2-2xy+y^2)=1. Thus (x-y)(x-y)^2 = 1=> (x-y)^3 = 1. So x-y=1. Also x+y=3. Thus x=2, y=1.

[canaldocentdenjosepmarti]

If you add the two equations you have (x+y)^3=27
So x+y = 3

If you substract the second equation from the first one you have
(x-y)^3 = 1
So x-y = 1

Then x=2 ; y=1

[sawyerw]

I used your 2nd method. One quibble is its much easier to substitute x=3-y in 2nd eqn, for obvious reason you only have to square 3-y rather than cube it.

[turkishwagnerian]

1. denklem + 2. denklem= 27 (küp kök) x+y=3
1. denklem - 2. denklem= 1 (küp kök) x-y=1 x=2 y=1

[divyanshsrivastava]

2 and 1 just add and subtract both eqn to get two eqns

[inyks]

2つ目の方法ですが上の式から下の式を引くと
x^3+3xy^2-3x^2*y-y^3=1
⇔(x-y)^3=1
⇔x-y=1
が出てくるので
x+y=3
x-y=1
の連立方程式になるので楽です。

[nosnibor]

YES YES !!! I forgot about the "Lets try a solution method". We Engineers call it the "suck it and see method". I tried lots of ways purely analytically.

[shmuelzehavi]

If you don't restrict yourself to real numbers, then you'l get the following 9 complex solutions:
x = ( 3 exp⁡(i 2mπ/3) + exp⁡(i 2nπ/3) ) / 2
y = ( 3 exp⁡(i 2mπ/3) - exp⁡(i 2nπ/3) ) / 2
where: m, n ∈ {0, 1, 2}
Those solutions include the solution presented in this clip (for m=n=0).

[carloshuertas]

Another great explanation, SyberMath! I actually solve for x and y in my head. Thanks a lot!

[YoutubeModeratorsSuckMyBalls]

Actually, we can reduce system to another simpler system:
X-y= 1 and x+y=3
So x=2 and y =1

[Drk]

Piece of cake! Cube of sums and differences. Adding/resting eqns, the system becomes: (x+y)^3=27---> x+y=3; (x-y)^3=1---> x-y=1. The new linear system gives us x=2; y=1. I hadn't thought the homogeneus method, interesting too. Always learning in this channel :)

[mathswan]

Adding the two equation gives (x+y)^3=27 hence x+y=3 sub into original equation we have x=2 and y=1

[jarikosonen]

To use the y=kx substitution the m+n of x^n*y^m must be same in all terms..

[KoiMorris]

What I did was the same as your second method up until the point you got x + y = 3. I realized that the answer had to be (x, y)= (1, 2) or (2, 1). I plugged in each answer into each of the original equations to see which one works in both. Only the latter answer works.

[krishnanadityan]

Add the two equations. The LHS is (x+y)^3.
Subtract the two and we get (x-y) ^3.
Solve the two equations to get x and y.

[chrisng]

Before I watch the video, here's my solution

Notice that x3 + 3xy2 + y3 + 3x2y = (x+y)3, and that 13+14 = 27. We can guess (x+y) therefore has to be 3, and we can guess that (x+y) = (1+2) or (2+1). We can see from the original 2 equations that x3 + 3xy2 = 13, and if we plug in (1)3 + 3(1)(2)2, we indeed get 13, therefore x=1, y=2

[moeberry]

By the way just for fun x is also (5+sqrt(3)i)/4 and also it’s conjugate (5-sqrt(3)i)/4 and y is those values subtracted from 3.

[TiahraThankyew]

I factored out the y in the second equation. Then noticed as 13 is prime either y is 1 or 13 and the inside of the bracket must be the other. Y=13 won't work as the second equation will be larger than 13. Sub y=1 into second equation and you will get x=+-2 only +2 works with y=1 in equation one

[ertugrulberatbilici]

Vaaay hocam Türkmüşsünüz siz ya 😅 Matematik ve bilim aşığı bir fen lisesi öğrencisi olarak selamlar. Çözümleriniz yeni bakış açıları kazandırıyor teşekkür ederiz.