Factoring a cubic polynomial

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#ChallengingMathProblems #Putnam #polynomialequations

SAME PROBLEMS FROM DIFFERENT PERSPECTIVES
Solving A Cubic Equation in Two Variables:
A Polynomial Equation Adapted from Putnam Exam 2006

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[SyberMath]

SAME PROBLEMS FROM DIFFERENT PERSPECTIVES
Solving A Cubic Equation in Two Variables:
A Polynomial Equation Adapted from Putnam Exam 2006

[SyberMath]

The latest videos were kind of longer. Hopefully this makes up for that. An easyish factoring problem. The solution is cool! Enjoy! 😊

[mathopediamathexplorer]

Well u can directly use the identity
a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2+ b^2 + c^2 - ab - bc - ca)
First write the expression as
X^3 + y^3 + (-1)^3 - (3xy*-1)
Then do the needful ...

[backgammonmaster]

I think you made an error at 4:10 with difference of two cubes identity, the signs should all be positive in the second part . you wrote minus (X+Y) . check it out . GREAT WORK ALL THE SAME !

[kennethstevenson]

The problem that might yield a cubic expression of this type is appropriate since our world is three dimensional and in some cases four dimensional. It might be easier to understand if it was used to solve a problem. The factoring does not seem to make the original expression easier to use.

[billmorrigan]

You've made a mistake: x^3 - 1 = (x - 1)(x^2 + x + 1), and not (x - 1)(x^2 - x + 1). Therefore the correct answer to your problem is (x + y - 1)(x^2 + y^2 - xy + x + y + 1).

[raffaelevalente]

It is incorrect
$x^3+3 x y+y^3-1=(x+y-1) \left(x^2-x y+y^2+x+y+1\right)$

[BCS-IshtiyakAhmadKhan]

Same question was in my jee practice sheet but instead of x and y it was sinx and cosx

[ashishpradhan]

Yes these take a little bit extra time. But it's fine. We enjoy a lot.Waiting for next geometry puzzle!!?😊😊😊😀😀😀

[tomasbobadilla]

Is (x+y-1)( (x+y) ^2 + (x+y) +1) because is deference of cubes brother.