Question 1090594
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If x^3 - y^3 = 485 and x-y = 5, find the numerical value of x + y. 
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Recall factoring (short cut)  x^3 - y^3 = (x-y)*(x*2 + xy + y^2).

Then you have  485 = x^3 - y^3 = (x-y)*(x^2 + xy + y^2) =  (replace  x-y  by 5)  = 5*(x^2 + xy + y^2),  
which implies  (after canceling the factor 5)

{{{485/5}}} = (x^2 + xy + y^2).


So, you have this equation 

x^2 + xy + y^2 = 97.   (1)


From x-y = 5 express y = x-5 and substitute it for y into equation (1). You will get

x^2 + x*(x-5) + (x-5)^2 = 97.

Simplify it step by step and then solve for x:

x^2 + x^2 - 5x + x^2 - 10x + 25 = 97,

3x^2 - 15x - 72 = 0,

x^2 - 5x - 24 = 0.     (2)

Factor left side to get an equivalent equation

(x-8)*(x+3) = 0.


The roots are  x = 8 and x = -3.


<U>Answer</U>.  One solution is x= 8,  y= 3.   Then x + y = 11.

         The other solution is x= -3,  y= -8.   Then x + y = -11.
</pre>


<U>Notice</U>. &nbsp;What I actually did in the course of solution, &nbsp;was reducing the given equation of the degree &nbsp;3&nbsp; to the quadratic equation
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;using given information.



Solved.