<s>[INST] Show that there is only one real root of the equation $1 + a^2 + ax - x^3$ I have to show that for every $a>0$ the equation $1 + a^2 + ax - x^3 = 0$, has exactly one solution. I have made the graph of the function $f(x) = 1 + a^2 + ax - x^3$ on desmos and I can clearly see that. But, how can I prove it explicity?
Thanks.
 [/INST] Write $f(x)=x^3-ax-a^2-1.$ Then we have that $f'(x)=3x^2-a.$ Hence the function has its local extrema at $$x=\pm\sqrt {\frac a3}.$$ It follows that there is exactly one real root if the relative extrema have equal sign. Hence we compute $$f\left(\sqrt {\frac a3}\right)f\left(-\sqrt {\frac a3}\right)$$ and confirm that it is positive. This gives $$\left[a^2+1 +a\sqrt {\frac a3}-\left(\sqrt {\frac a3}\right)^3\right]\left[a^2+1-\left(a\sqrt {\frac a3}-\left(\sqrt {\frac a3}\right)^3\right)\right]=(a^2+1)^2-\left[a\sqrt {\frac a3}-\left(\sqrt {\frac a3}\right)^3\right]^2=(a^2+1)^2-\frac a3\left(a-\frac a3\right)^2=a^4-\frac{4}{27}a^3+2a^2+1=\left(a^2+\frac{2a}{27}\right)^2+\left(2-\frac{4}{27^2}\right)a^2+1,$$ which is clearly positive for all $a.$
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