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- Feb 14, 2012

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I hope to gain some insights on how to solve this system of equations because I tried it many times to use trigonometric approach but to no avail...

Problem:

Solve the system of equations

$x^4-y^2-xy=4-\sqrt{15}$

$x^3+y^3-3x=5\sqrt{5}$

Attempt:

At first glance, $x^3-3x$ suggests the substitution of $x=\tan p$ so that $x^3-3x=\tan^3 p-3\tan p=(3\tan^2 p-1)(\tan 3p)$

and the second given equation can be rewritten as below, if we also let $y=\tan q$,

$x^3-3x=5\sqrt{5}-y^3$

$(3\tan^2 p-1)(\tan 3p)=5\sqrt{5}-\tan^3 q$

$\tan^3 q=5\sqrt{5}-(3\tan^2 p-1)(\tan 3p)$

and the first equation becomes

$x^4-y^2-xy=4-\sqrt{15}$

$\tan^4 p-\tan^2 q-\tan p \tan q=4-\sqrt{15}$

$\tan^2 q+\tan p \tan q+4-\sqrt{15}-\tan^4 p=0$

Even if I solve the equation above for $\tan q$ by the quadratic formula, I can see that this is of a futile attempt...

Could someone please help me to solve this very hard problem for me? Thanks in advance.