Solving a Cubic System

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Overcomplicated it at the beginning, since you just needed eq1 + eq2, eq1 + eq3, eq2 + eq3. After that it was fine. Although I would have left the solutions as x^3 = c, then you would have the complex-valued solutions as well, just by finding all three cube roots for each (although you have to match them up carefully, take pairs of almost-complex-conjugates)

XJWill
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To remove the cube roots from the denominator, multiply the numerator and denominator by the cube root of 169.

bobbyheffley