Solving a cubic system

This video is about solving a cubic system of equations. Surprising solution at the end.
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[aahaanchawla]

Just one doubt. We will have to take either x1, y1 and z1 or x2, y2 and z2 right? Or can we take any combination of the six values?

[vizirudominic]

So I have a question: So the answer is Ans(Sol.)={(x=5/cuberoot13, y=2/cuberoot13, z=6/cuberoot13); (x=1, y=-2, z=2)} ???

[leecherlarry]

too many solutions, compi found 18 solutions (incl. complex), yikes! haha:
*Solve[{x^3 + y^3 - z^3 - x y z + 11 == 0, x^3 - y^3 + z^3 - x y z - 21 == 0, -x^3 + y^3 + z^3 - x y z + 3 == 0}, {x, y, z}]*

[mxsjncv]

I looked at it as if the first equation is a+b-c-d=-11 and the other equations accordingly. Found that the infinite solutions to this linear-system can be written as ( t-7, t-16, t, t-12) . now we know that (t-12)^3 = t(t-7)(t-16).

[akinsdecaprio]

Mistake in one of your steps. Cross check it

[aashsyed]

/-\ THE FANCY VERSION OF "A"