<s>[INST] 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)
 [/INST] $u^{12}+2u^9+3u^8+4u^6+4u^5+3u^4+8u^3+4u^2+2u+17=u^8(u^4+2u+3)+(x+1)^2(4x^4-4x^3+7x^2-2x+1)+2(x+1)+14$
$u^4+2u+3 \ge 2u^2+2u+2 \ge \dfrac{3}{2}$
$4x^4-4x^3+7x^2-2x+1=x^2(2x-1)^2+6(x-\dfrac{1}{6})^2+\dfrac{5}{6} > \dfrac{5}{6}$
$(x+1)^2(4x^4-4x^3+7x^2-2x+1)+2(x+1)+14>\dfrac{5}{6}(x+1)^2+2(x+1)+14=\dfrac{5}{6}(x+1+\dfrac{6}{5})^2+12\dfrac{4}{5}>12\dfrac{4}{5}$
$u^{12}+2u^9+3u^8+4u^6+4u^5+3u^4+8u^3+4u^2+2u+17>12.8$
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