<s>[INST] Given $a^4+8b=4(a^3-1)-16\sqrt 3$ and $b^4+8a=4(b^3-1)+16\sqrt 3$, find $a^4+b^4$ Given $a^4+8b=4(a^3-1)-16\sqrt 3$ and $b^4+8a=4(b^3-1)+16\sqrt 3$, find $a^4+b^4$
I tried adding and subtracting both equations, but didn't get anywhere. Would appreciated any ideas. Thanks!
 [/INST] Adding both equation gives:
$$a^4+b^4+8(a+b)=4(a^3+b^3)-8$$
i.e: $$(a^2-2a)^2-4(a^2-2a)+b^4-4b^3+8b+8=0$$
Consider this equation is a quadratic equation with variable $a^2-2a$
So the discriminant is:$(-2)^2-(b^4-4b^3+8b+8)=-b^4+4b^3-8b-4=-(b^2-2b-2)^2$
Then:$b^2-2b-2=0$, so $b=1-\sqrt{3}$ or $b=1+\sqrt{3}$ then you can easily find $a=1+\sqrt{3}$ and $b=1-\sqrt{3}$ is only solution which is satisfied. Then $a^4+b^4=56$
</s>