I just made a list of cubes of 1 to 10, ie, 1 to 1000. Cubed x must be greater than cubed y, so x>=5 (125 for x=5 vs 64 for x=4). The difference between adjacent cubes (ie, smallest distance) must also be less than 91, so y
It's called "teaching." "Teaching" takes longer than the learned doing. There's no indication you've solved this. But feel free to be snarky and irrelevant.
At start you have to Say, Solving on integer numbers
exactly, there are infinite solutions
Observe that 91 is sum of cubes of 4 and 3 I.e 91= 64 + 27. Hence, answer is x = 4 and y = -3
x-y is not always less than x x+x y +y y. Let x=1 and y=-1. Then, x-y=2 but x x+x y +y y=1-1+1=1
x^3 = 91 - y^3
x = the cube root of (91 - y^3)
I just made a list of cubes of 1 to 10, ie, 1 to 1000.
Cubed x must be greater than cubed y, so x>=5 (125 for x=5 vs 64 for x=4).
The difference between adjacent cubes (ie, smallest distance) must also be less than 91, so y
Очень уж длинное решение,так явно видно,что х=6 а у=5;
Отрицательных ответов не надо для куба-неверно !!!
6³-5³= 91
216-125=91
91=91
100-9=91
What had to be solved in whole numbers?
THANKS PROFESOR !!!, VERY INTERESTING !!!!
You are welcome! Thank you very much!!
At no time did you indicate that x and y are integers. Your solution is incomplete until you check for other real, and indeed, complex solutions.
45.5 x 2 also = 91
There’s actually an infinite number of solutions. This probably would map as an interesting 3d object
@@ccudmore
Two dimensional instead.
NONSENSE: one equation with two unknowns, you need as many equations as degrees of freedom.
as I understand the only integer solutions required
How you assume that product of x and y is positive in the beginning?
Sir, because x.y=30 it is not posible for -4,3 and -3,4?
a cubic with 4 solutions. cool. (a double cubic i suppose)
X=6, Y=5.
X^3 - Y^3 = 91
x-y
多分誤りですね...反例も容易に見つけられてしまいますし...
(x,y) ∈ R²
x-y < x²+xy+y²
⇔x²+xy+y²-x+y > 0 ···①
だから
f(x,y):=x²+xy+y²-x+y=0
のグラフを書いてみる。
すると明らかに①を満たさない（x,y）∈Z² が存在することがわかる。
またf(x,y)は楕円の方程式3x²+y²=2を反時計回りに45度だけ回転させた後、x軸方向に1,y軸方向に-1だけ平行移動したもの。
wrong. true is 4（x^2+xy+y^2）≥（x-y）^2，that is （x+y）^2≥0。so only （1，91）and （7，13）can be possible solution。
@@windyyw なぜx-y>0といえるの？
X=5&y=4
x=6, y=5
x=6 and y=5.
1x91=91
6³-5³=91
Die Japanner snap niet niet veel van. Oneindig veel oplossingen. Niet de eerste keer dat hij in de fout gaat.
(6;5) (5;6)
You based your solution on assumptions and trial and error
That isn’t logical nor acceptable
Không có điều kiện x,y là số nguyên nên suy ra cách giải này sai.
Đáp án bài này là vô số nghiệm
Use soroban
废话太多了
Почти 19 минут расписывать то,что решается за 4 минуты? Вы издеваетесь над нами?
It's called "teaching."
"Teaching" takes longer than the learned doing.
There's no indication you've solved this.
But feel free to be snarky and irrelevant.
Everything is trivial if you know the solution. However, that's not the point.