Question 1210378
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x+y = 10
(x+y)^3 = 10^3
x^3+3x^2y+3xy^2+y^3 = 1000 .......... binomial theorem
(x^3+y^3) + (3x^2y+3xy^2) = 1000
(x^3+y^3) + 3xy(x+y) = 1000
300+3xy(10) = 1000  ............. plug in x+y=10 and x^3+y^3=300
300+30xy = 1000
30xy = 1000-300
30xy = 700
xy = 700/30
xy = 70/3



x+y = 10
x(x+y) = 10x ...... multiplying both sides by x
x^2+xy = 10x
x^2+70/3 = 10x ........... plug in xy = 70/3 found earlier
3x^2+70 = 30x ........... multiply both sides by the LCD 3
3x^2-30x+70 = 0
I'll let the student finish up.
I recommend using the quadratic formula.


You should get two distinct real number solutions for x. 
Once determining x, you can determine the paired value of y.


Due to symmetry, if (a,b) is a solution then (b,a) is the other solution.
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