Question

(х-3)^4 + (x-6)^4=16 

Answer 1

$x-3=a\ x-6=a-3\ \ a^2+(a-3)^2=16\ 2a^4-12a^3+54a^2-108a+65=0\ (a^4-6a^3+9a^2)+a^4-6a^3+45a^2-108a+65=0\ (2a^4-12a^3+18a)+36a^2-108a+65=0\ 2(a^2-3a)^2+36(a^2-3a)+65=0\ a^2-3a=t\ \ 2t^2+36t+65=0\ D=36^2-4*2*65=sqrt{776}\ t_{1}=frac{36+sqrt{776}}{4}\ t_{2}=frac{36-sqrt{776}}{4}\ \ a^2-3a=frac{-36+sqrt{776}}{4} \ \ a=frac{3-/+sqrt{2sqrt{194}-27}}{2}\ x-3=frac{3-/+sqrt{2sqrt{194}-27}}{2}\ x=frac{3-/+sqrt{2sqrt{194}-27}}{2}+3$