An Interesting Cubic Equation

[@shogun6943]

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

[@elecbertelecbert-l5e]

You are genius super genius a real human with hope all my educators said humans must be genius but if they are not they are animals because animals can not solve mathematics but actually all my school solve nothing and fail in the finally exams you are so brilliant keep going 🎉🎉

[@KennethChile]

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?

[@bosorot]

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.

[@giuseppemalaguti435]

...(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

[@arekkrolak6320]

This is becoming a joke. Why discard two solutions out of three?

[@bosorot]

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

[@XJWill1]

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 }

[@scottleung9587]

I got -2 as my answer.

[@joni123451000]

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