An Interesting Cubic Equation
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- Опубліковано 9 січ 2025
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a^3-b^3-3ab=1
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a^3 -b^3 -3ab=1
a^3 + ( -b)^3 + (-1)^3 = 3(a)(-b)(-1)
Therefore a + (-b) + (-1) = 0
Hence a - b = 1
Because when a^3 + b^3 + c^
3 = 3abc then a + b + c = 0
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If you graph in desmos or Wolfram x^3-y^3-3ab=1 you obtain the line y=x-1, so x-y=1 is consistent with your first solution, but not the second x-y=-2, why?
To better visualize on demos , use a slider instead of number 3. move it around and you will see that only number 3 will give you this special relationship that made a-b a constant . For a=-1 and b =1 . it is only a dot / not a line . It will not show a dot on graph.
...(a-b)=(1+3ab)/((a-b)^2+3ab)...1/(a-b)=((a-b)^2/(1+3ab))+1...(1-a+b)/(a-b)=(a-b-1)(a-b+1)/(1+3ab)....quindi una soluzione è a-b-1=0,a-b=1
This is becoming a joke. Why discard two solutions out of three?
if you mean by 4:20 . he did not discard 2 solutions . he just stated that the method to use quadratic formula is not working , need other way to solve it
The explanation left a lot to be desired, and the solution was all over the place. But he did mention at the beginning that a and b are limited to real values. Here is my solution using that assumption.
a^3 - b^3 - 3*a*b - 1 = 0
(a - b - 1)*( a^2 + (b+1)*a + b^2 - b + 1 ) = 0
So this expression is true if
a - b = 1
OR it is true if
a^2 + (b+1)*a + b^2 - b + 1 = 0
This is a quadratic equation in variable a, and we require a to be real-valued. Therefore, the determinant must be non-negative:
(b+1)^2 - 4*1*(b^2 - b + 1) >= 0
- 3 * (b - 1)^2 >=0
b = 1
The determinant is only non-negative for one single value of b. And letting b = 1 in the previous equation results in
(a + 1)^2 = 0
so a = -1
Therefore, the complete solution set of the given equation, ASSUMING a and b ARE REAL-VALUED, is:
{ [a - b = 1] , [ a = -1 , b = 1] }
But the problem asked for a - b, so the complete set of possible values for a - b, given a and b are real-valued, is:
{ a - b = 1 , a - b = -2 }
I got -2 as my answer.
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