<s>[INST] Roots to the quartic equation, $(x+1)^2+(x+2)^3+(x+3)^4=2$ Solving with Mathematica gives me the four roots, $$x=-4,-2,\dfrac{-7\pm\sqrt5}{2}$$ Is there some trick to solving this that doesn't involve expanding and/or factoring by grouping?
 [/INST] Set $y=x+2$. The equation rewrites as:
$$(y-1)^2+y^3+(y+1)^3==2\iff y(y^3+5y^2+7y^+2)=0$$
The second factor has an integer root: $-2$, hence the factorisation:
$$y^3+5y^2+7y +2=(y+2)(y^2+3y+1).$$
Finally, the roots (in $y$) are: $0, -2, \dfrac{-3\pm\sqrt 5}{2} $, whence the roots of the initial equation:
$$ \Bigl\{ -2,-4, \frac{-7\pm\sqrt 5}{2}\Bigr\}.$$
</s>