a^3+b^3=10,a^2+b^2=7,a+b=x? a,b£R
given: a^3 + b^3 = 10 and a^2 + b^2 = 7
and x = a+b
(a+b)^2 = a^2 + 2ab + b^2
x^2 = 7 + 2ab
ab = (x^2 - 7)/2
(a + b)^3 = a^3 + 3a^2b + 3 a b^2 + b^3
x^3 = a^3 + b^3 + 3ab(a + b)
x^3 = 10 + 3((x^2-7)/2) (x)
x^3 = 10 + 3x(x^2 - 7)/2
2x^3 = 20 + 3x^3 - 21x
x^3 - 21x + 20 = 0
by quick inspection, x = 1 , so x-1 is a factor, by division
x^3 - 21x + 20 = (x-1)(x^2 + x - 20) = 0
(x-1)(x+5)(x-4) = 0
x = 1, -5, 4
If you want to find a and b, you have 3 cases:
case 1: x = 4
then a+b = 4, ab = (x^2 - 7)/2 = 9/2 or b = 9/2a
a + 9/(2a) = 4
2a^2 + 9 = 8a
2a^2 - 8a + 9 = 0
a = (8 ± √-8)/4 = (8 ± 2√2 i)/4
= (4 ± √2 i)/2
if a = (4 + √2 i)/2 , then b = 4 - (4 + √2 i)/2 = (8 - 4 - √2 i)/2 = (4 - √2 i)/2
b = (4 - √2 i)/2
repeat for the other two values of x