SOLUTION: 1) If x + y = 1 and x³ + y³ = 3, find (1/x³) + (1/y³) 2) If x^(x¹⁶) = 16, find x²⁰

        Part  (a)

Start from

    x + y = 1.


Raise it in degree 3.  You will get

    1 = (x+y)^3 = x^3 + 3x^2*y + 3x*y^2 + y^3 = (x^3 + y^3) + 3xy(x+y).


Replace here x^3 + y^3  by  3 (since it is given),  and replace  x+y  by  1.  You will get

    1 = 3 + 3xy ,

    3xy = -2,

     xy = .


Now    +  =  =  =  = .


ANSWER.  At given condition,   +  = .

        Part  (b)

Obviously, the value of x should be greater than 1.


One such value is easy to guess:  it is  x = 2^(1/4) =  = 1.189207115,  approximately.


Indeed,  (2^(1/4))^16 =  = 16  and,  therefore,

    x^(x^16) = (2^(1/4))^16 =  = 16,


so this value of x,  x = 2^(1/4)  is the solution to the given equation  x^(x¹⁶) = 16.


From the other side hand,  this function, f(x) = x^(x¹⁶)  is a monotonically increasing function 
in the domain x > 1, so, there is no other solution to the given equation in this domain.



        Thus, for positive real numbers, the only solution 

            to the given equation  x^(x¹⁶) = 16  

        is  x = 2^(1/4) =  = 1.189207115  (approximately).



CHECK.   = 16;  therefore,  x^(x¹⁶) = x^16 = 1.189207115^16 = 16,  which is correct.


Now,   = (2^(1/4))^20 = 2^5 = 32.


ANSWER.  Under given condition,   = 32.