Let $a$ and $b$ be real numbers such that $a^3 + 3ab^2 = 679$ and $a^3 - 3ab^2 = 615.$ Find $a - b.$
+120 a^3 + 3ab^2 = 679
a^3 - 3ab^2 = 615
We can subtract both of the equations :
(a^3 + 3ab^2 ) - ( a^3 - 3ab^2 ) = 679 - 615
= 6ab^2 = 64
= ab^2 = 64/6
Now we need to get rid of the fraction part of 64/6, thus a = 1/6
If a = 1/6 then b = ± 8
Now we subtract, we have 2 possibilities :
1/6 - (-8 ) = 8 + 1/6
Or
1/6 -8 = -47 / 6
