<s>[INST] Solve $x^4+3x+20=0$ by Ferrari's method Comparing the equation 
$$x^4+3x+20=0$$
With the equation 
$$(x^2+\lambda)^2-(mx+n)^2=0$$
we get    
$m^2=2\lambda,$
$-2mn=3,$
$n^2=\lambda^2-20$   
Now, $4m^2n^2=9\Rightarrow 4(2\lambda)(\lambda^2-20)=9\Rightarrow 8\lambda^3-160\lambda-9=0$.   
How can I find easily the values of $\lambda$ from the above equation.
Please suggest me.
 [/INST] HINT: Use binom expansion 
$$(x+y)^3=x^3+3x^2y+3xy^2+y^3$$
Reorder the equation as:
$$(x+y)^3-3xy(x+y)-(x^3+y^3)=0$$
$$8\lambda^3-160\lambda-9=0$$
 $$\lambda^3-20\lambda-\frac{9}{8}=0$$   
Define:
$x+y=\lambda$,
then solve the equation with 2 unknowns 
$$3xy=20$$
$$x^3+y^3=\frac{9}{8}$$
Let me know if you cannot get forward from there.
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