GIF Will Trouble You

Let's define a function $f\left( x \right) = x^5 - x^3 + x$ and $f'\left( x \right) = 5x^4 - 3x^2 + 1$ . The equation $f'(x) = 0$ has no real solutions and as $f'(0) = 1$ , then $f'(x)$ is strictly positive, and thus $f(x)$ is stricty increasing. Therefore, the equation $f(a) = 2$ has at most one solution for $a$ .

Now, let's notice that $f(1) = 1 < 2$ and $f(2) = 26 > 2$ . By the Intermediate Value Theorem, there exist at least one solution to $f(a) = 2$ in the interval $\left( 1 , 2 \right)$ . It's clear now that $a$ is unique and lies in $\left( 1 , 2 \right)$ , therefore $a^6$ lies in $\left( 1 , 64 \right)$ . Evaluating $f(x)$ at some specific points we can reduce the interval as in the bisection method: $f\left( {\sqrt[6]{2}} \right) \approx 1.4900$ $f\left( {\sqrt[6]{3}} \right) \approx 1.9669$ $f\left( {\sqrt[6]{4}} \right) \approx 2.4347$ Therefore, $a$ lies in $\left( {\sqrt[6]{3},\sqrt[6]{4}} \right)$ , and $a^6$ lies in $\left(3 , 4\right)$ . In conclusion: $\left[ {a^6 } \right] = \boxed{3}$ .

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