How Can I Solve x^2-xy+y^2+x+y+1=0 for My Math Project?

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To solve the equation x^2 - xy + y^2 + x + y + 1 = 0 for a math project, it can be treated as a quadratic in y. The quadratic formula can be applied after rearranging the equation to identify coefficients. The discriminant of the quadratic will determine if there are real solutions; a negative discriminant indicates no graph, while a positive one provides a formula for y. An alternative approach involves expanding the expression to explore potential insights. Understanding these methods will aid in plotting the graph effectively.
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I need help with this Problem x^2-xy+y^2+x+y+1=0. It is part of a math project I am doing to plot a graph. The original problem was x^3+3xy+y^3=1 can factor and it plots a line at a 45 degree angle. The first part of the factorization is x+y-1=0. The only thing I can figure to do with the problem is to take the +1 and move it to the otherside of the = and make it -1 so I have a point to work with other than 0. The only number I have found that will equal -1 is to have both x and y as -1. Any help would be gladly received. Thanks for your time.

P.S. I am only in Intermediate Algerbra so please make it so I can understand.
 
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You started with your original equation:
x3+3xy+y3 - 1 = 0
and factored it to get:

(x + y - 1)(x2 -xy+y2+x+y+1) = 0

and have used the first factor to get the straight line. So your remaining problem is to see if the other factor gives you anything:

x2- xy+y2+x+y+1 = 0

This is a quadratic equation in y which you can solve by the quadratic formula. So collect terms on powers of y and look at the discriminant of that quadratic equation. If the discriminant is negative you get no graph. If it is positive the quadratic formula will give you a formula for y.
 
If you're confused with what to do with the quadratic, as an alternative, try expanding:

\frac{1}{2}\left((x-y)^2+(x+1)^2+(y+1)^2\right)

and see if you can deduce anything from that.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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