How to Solve a Cubic System by Eliminating Variables?

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To solve the cubic system defined by the equations x³+y³=1 and x²y+2xy²+y³=2, one approach is to isolate y in the first equation and substitute it into the second, resulting in an equation solely in terms of x. Alternatively, the second equation can be rearranged into a quadratic form in x, allowing the use of the quadratic formula to find solutions. Another method involves expanding (x+y)³ and substituting known values, simplifying the process. By letting r=x/y and dividing both equations by y³, a cubic equation in r can be derived, which can be solved by finding a root through guessing or graphing. Once a root is identified, it can be used to reduce the cubic equation to a quadratic for further solutions.
Dorotea
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x³+y³=1
x²y+2xy²+y³=2
 
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You could try solving the first equation for y and substituting it int the second...that will give you an equation in terms of x only.
 
Or you could solve for x in the 2nd equation, which is quadratic in x, by using the quadratic formula.
yx^2 + 2y^2 x + y^3 -2 = 0
You will of course get two equations for x.
 
You could expand (x+y)3 and then substitute the values for the two expressions that you have. That would be a lot simpler.
 
Notice that both equations only have terms of degree 3 in both variables. So one thing you could do is let r=x/y, and divide both equations by y^3. Each left-hand side will then depend only on r, and each right-hand side only on y^3. Eliminating y^3 will give you a cubic equation in r. Solve this by guessing a root, or by graphing to get a root. Once you know one root you can reduce it to a quadratic to get the other two.
 

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