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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)
= (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.
Answer. One solution is x= 8, y= 3. Then x + y = 11.
The other solution is x= -3, y= -8. Then x + y = -11.
Notice. What I actually did in the course of solution, was reducing the given equation of the degree 3 to the quadratic equation
using given information.
Solved.