Maximum value

Here X+Y=1 so X^3+Y^3 +3XY(X+Y)= 1 => X^3+Y^3+3XY=1 [ given, X+Y=1] => X^3+Y^3= 1-3XY and so alpha= (X^3+1) (Y^3+1)= X^3+Y^3+X^3 Y^3+1 = X^3 Y^3+1+1-3XY = X^3 Y^3-3XY+2 now let XY=D so we can write alpha = D^3-3D+2 and now differentiate alpha with respect to D. and we get 3D^2-3 which must be equal to 0 for getting the maximum value of the above expression. and then D=XY= +/- 1 and given X+Y=1 and we get X-Y = square root 5 by putting value of XY =-1 cause if we putting XY=+1, X-Y will become an imaginary number which is invalid. Then we get the value of X and Y as (1+Sq. root 5)/2 and (1-sq. root 5)/2 respectively. and by putting these values we get the maximum value of alpha =4 Answer: 4

Start by writing y in terms of x

$( (1 - x)^{3} + 1)( x^{3} + 1)$

Expand and differentiate it with respect to x

$\frac{d}{dx} (-x^{6} + 3 x^{5} - 3 x^{4} + x^{3} + 3 x^{2} - 3 x + 2)$

Find the roots of the resultant derivative; this is done by equating it to zero and solving for x

$- 6 x^{5}+15 x^{4} - 12 x^{3} + 3 x^{2} + 6 x - 3 = 0$

Once this is done you will have 3 real roots (and 2 imaginary/complex roots which we will neglect) The real roots would be the following: $\frac{1}{2}$ , $\frac{1}{2} - \frac{\sqrt{5}}{2}, \frac{1}{2} + \frac{\sqrt{5}}{2}$ )

Finally plug any of the real roots , with an exception to $\frac{1}{2}$ , back into the original function and you would get the maximum value

E.g. $((1 - \frac{1}{2} + \frac{\sqrt{5}}{2})^{3} + 1)( (\frac{1}{2} - \frac{\sqrt{5}}{2})^{3} + 1) = \boxed{4}$