A Cubic System of Equations

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without watching
8(eq2) - (eq1): 12x³ - 19x²y + 8xy² - y³ = 0
set y = kx
x³(12 - 19k + 8k² - k³) = 0
by eq2 x cannot be zero
k³ - 8k² + 19k - 12 = 0
the coefficients sum to 0; extract (k - 1) and factor the remaining quadratic
(k - 1)(k² - 7k + 12) = 0
(k - 1)(k - 3)(k - 4) = 0
substitute possible values of k into eq2:
x³ = 1 or 5x³ = 1 or 10x³ = 1
(pretty much done at this point)

coreyyanofsky
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just done by the evaluation of coefficients

broytingaravsol
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In the first method, how do you know y=mx captures all the possible solutions, other than because (in this narrow case) it's a cubic so there are three ?

neuralwarp
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I have made system of eqn, please solve this for real solutions :
(a+b+c+d)/4 = 4th root
ln(ln(ln(a^b^c^2))) =

itzerr