Solve $t^4+4 t^3+6 t^2+4 t-32 t^{1/4}+1 = -16 $ I'm trying to solve the following equation: $$(t+1)^4 - 32 t^{\frac{1}{4}}=-16 $$ where t $\geq 0$,
which is equivalent to $$t^4+4 t^3+6 t^2+4 t-32 t^{\frac{1}{4}}+1 = -16 $$
Wolfram Alpha tells that the equation is equivalent to $$t+1=2t^{\frac{1}{4}} $$ if we assume that t>0. How to prove this? The hard part is how to factorise... Is there another solution? (BTW, without to obtaina 3-rd degree equation and to use Cardano)
 A: Well, let me post a partial answer. Let's make the substitution $t=u^4$ for $u\ge0,$ which gives us the equation $$(u^4+1)^4-32u=-16,$$ or $$u^{16}+4u^{12}+6u^8+4u^4-32u+1=-16.$$ We wish to show that this is equivalent to $$u^4+1=2u.$$ That is, we must show that for $u>0,$ we have $$u^{16}+4u^{12}+6u^8+4u^4-32u+17=0\tag{1}$$ if and only if $$u^4-2u+1=0.\tag{2}$$
Indeed, polynomial long division shows us that the polynomial in $(1)$ can be written as $$(u^4-2u+1)(u^{12}+2u^9+3u^8+4u^6+4u^5+3u^4+8u^3+4u^2+2u+17).$$ The tricky part is in showing that that 12th degree polynomial factor has no zeroes (which will complete the proof). Perhaps someone else will come along with a nice method of doing so.
A: $u^{12}+2u^9+3u^8+4u^6+4u^5+3u^4+8u^3+4u^2+2u+17=u^8(u^4+2u+3)+(x+1)^2(4x^4-4x^3+7x^2-2x+1)+2(x+1)+14$
$u^4+2u+3 \ge 2u^2+2u+2 \ge \dfrac{3}{2}$
$4x^4-4x^3+7x^2-2x+1=x^2(2x-1)^2+6(x-\dfrac{1}{6})^2+\dfrac{5}{6} > \dfrac{5}{6}$
$(x+1)^2(4x^4-4x^3+7x^2-2x+1)+2(x+1)+14>\dfrac{5}{6}(x+1)^2+2(x+1)+14=\dfrac{5}{6}(x+1+\dfrac{6}{5})^2+12\dfrac{4}{5}>12\dfrac{4}{5}$
$u^{12}+2u^9+3u^8+4u^6+4u^5+3u^4+8u^3+4u^2+2u+17>12.8$
A: $f(t)=(t+1)^4-32t^{\frac 14}$
First, our domain is $[0,\infty)$ because otherwise $t^\frac14$ is not defined.
Derivative: $f'(t)=4(t+1)^3-8t^{-\frac 34}$
Derivative negative iff $(t+1)t^{1/4}<\sqrt[3]2$
so nonnegative iff $(t+1)t^{1/4}\ge\sqrt[3]2$
The equality occurs for $t=a$ where $0<a<1$. (Checkable by inspection since LHS is increasing and at $0$, LHS=$0$ and at $1$, LHS=$2$ so use intermediate value property to get LHS=$\sqrt[3]2$ for some $0<a<1$.)
Hence $f$ which is already negative at $0$ decreases until $a$ and increases thereafter. $f$ increases to $0$ at $1$ and since it's only increasing thereafter, there are no other solutions than $t=1$.
