Function #1

Finding the maximum of a sextic function sounds impossible.

However, since this is a special case, it is possible.

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Let $1-x=y$ and $xy=t$ .

Then $x+y=1$ .

Then the expression we're trying to get a maximum is $(x^3+1)(y^3+1)$ .

Simplify that and we get $(xy)^3+x^3+y^3+1$ .

Since $x^3+y^3=(x+y)^3-3xy(x+y)$ ,

The expression may be written as $(xy)^3+(x+y)^3-3xy(x+y)=t^3-3t+2$ .

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It is clear that $(x-y)^2\geq0$ .

$x^2-2xy+y^2\geq0 \\ x^2+2xy+y^2\geq4xy \\ (x+y)^2\geq4xy \\ \\ \therefore t=xy\leq\dfrac{(x+y)^2}{4}=\dfrac{1}{4}$

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$t^3-3t+2=t^3-3t-2+4=(t+1)^2(t-2)+4$

Note that since $t\leq\dfrac{1}{4}$ , $t-2$ is negative.

Since $(t+1)^2\geq0$ , $(t+1)^2(t-2)\geq0$ .

$\therefore\text{ }(t+1)^2(t-2)+4\geq4$ .

In order for the expression to have a maximum, $t$ must be equal to $-1$ so that $(t+1)^2(t-2)+4=4$ .

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Therefore the maximum of the given expression is $\boxed{\text{ 4 }}$ .