Anytime I see a system that contains x^3 +y^3 along with x+y, I think of cubing the linear expression, because it will contain the terms x^3 +y^3, along with another term with x+y as a factor. So (x+y)^3 = x^3 + 3x^2y + 3 xy^2 + y^3 = 8. Since x^3 +y^3 = 3, I am left with 3x^2 y + 3x y^2 = 5. That factors nicely to 3xy(x+y) = 5. But since x+y = 2, I get 6xy = 5, or xy = 5/6. The video got to the same place, but in a different way. Now combine this with x+y = 2 and we have a simple system that leads to the quadratic 6x^2+5 = 12x. The solution simplifies to x=1+sqrt(6)/6 or 1-sqrt(6)/6. The symmetry of the original system suggests that when x is the plus case, y is the minus case, and vice versa. And that is the way it works out.
Awesome! Thanks so much sir! I liked the video. So I hit the tumbs-up button and i got a reward. The thumbs-up button got bigger and blew up I mean there was confetti everywhere✓✓✓
Nice solution But i have different approach Let assume x and y solution of a quadratic equation x + y = 2 ...1 x^3 + y^3 = (x+y)(x^2 + y^2 -xy ) ...2 (x + y )^2 = 4 x^2 + y^2 = 4 - 2xy ...3 x^3 + y^3 = 3 Using ...1 ...2 ...3 3 = 2 (4 - 3xy) xy = 5/6 x + y = 2 So we have a simple quadratic with roots x and y - a^2 - (sum of roots )a + (product of roots ) = 0 So , a^2 -2a + 5/6 = 0 6a^2 - 12a + 5 =0 So x, y = 1 +/- ( root 6)/6 (By quadratic formula )
Let x = 1+p, y = 1-p. Then (1+p)^3+(1-p)^3=3 Simplyfying, 6p^2 = 1, p = +-1/sqrt6. (x, y) = (1+1/sqrt6, 1+1/sqrt6), (1-1/sqrt6, 1+1/sqrt6) P. S. remembered this method from one of your videos Upd: bruh
Anytime I see a system that contains x^3 +y^3 along with x+y, I think of cubing the linear expression, because it will contain the terms x^3 +y^3, along with another term with x+y as a factor. So (x+y)^3 = x^3 + 3x^2y + 3 xy^2 + y^3 = 8. Since x^3 +y^3 = 3, I am left with 3x^2 y + 3x y^2 = 5. That factors nicely to 3xy(x+y) = 5. But since x+y = 2, I get 6xy = 5, or xy = 5/6. The video got to the same place, but in a different way.
Now combine this with x+y = 2 and we have a simple system that leads to the quadratic 6x^2+5 = 12x. The solution simplifies to x=1+sqrt(6)/6 or 1-sqrt(6)/6. The symmetry of the original system suggests that when x is the plus case, y is the minus case, and vice versa. And that is the way it works out.
x^3+(2-x)^3=3
6x^2-12x+5=0
Trivial quadratic equation to solve.
exactly my approach, short and easy
Awesome! Thanks so much sir!
I liked the video. So I hit the tumbs-up button and i got a reward.
The thumbs-up button got bigger and blew up I mean there was confetti everywhere✓✓✓
Neat #2 solution. I solved it using the first method.
Great solution
Nice solution
But i have different approach
Let assume x and y solution of a quadratic equation
x + y = 2 ...1
x^3 + y^3 = (x+y)(x^2 + y^2 -xy ) ...2
(x + y )^2 = 4
x^2 + y^2 = 4 - 2xy ...3
x^3 + y^3 = 3
Using ...1 ...2 ...3
3 = 2 (4 - 3xy)
xy = 5/6 x + y = 2
So we have a simple quadratic with roots x and y -
a^2 - (sum of roots )a + (product of roots ) = 0
So , a^2 -2a + 5/6 = 0
6a^2 - 12a + 5 =0
So x, y = 1 +/- ( root 6)/6
(By quadratic formula )
Let x = 1+p, y = 1-p. Then (1+p)^3+(1-p)^3=3
Simplyfying, 6p^2 = 1, p = +-1/sqrt6.
(x, y) = (1+1/sqrt6, 1+1/sqrt6), (1-1/sqrt6, 1+1/sqrt6)
P. S. remembered this method from one of your videos
Upd: bruh