x^3+y^3 = (x+y)(x^2-xy+y^2)
6(x^2-xy+y^2) = 144
x^2-xy+y^2 = 24
y = 6-x, so
x^2-x(6-x)+(6-x)^2 = 24
x^2-6x+x^2+x^2-12x+36 = 24
3x^2-18x+12 = 0
3(x^2-6x+4) = 0
Look familiar?
Find one pair of real numbers, (x,y), such that x + y = 6 and x^3 + y^3 = 144.
1 answer