Is this a Diophantine or maybe, not

Writing the given equation as a cubic equation in $a$ we have $a^3+3ba+(b^3-1)=0$ . The discriminant of this equation is $-(b^3+1)^2$ which is less than or equal to $0$ . So for real roots of the equation, $b^3+1=0\implies b=-1\implies a^3-3a-2=0$ or $(a+1)^2(a-2)=0$ . So, $a=-1$ or $a=2$ . Hence $x=-2, y=1$ and $3x+30y=\boxed {24}$ .

Aliter -

$a^3+b^3+3ab=1\implies a^3+b^3+(-1)^3-3ab(-1)=0\implies \dfrac{1}{2}(a+b-1)[(a-b) ^2+(b+1)^2+(a+1)^2]=0$ . So $a+b=1$ or $a=b=-1\implies a+b=-2$ . Thus $x=-2,y=1$ and $3x+30y=\boxed {24}$

This can be rearranged as $a ^ 3$ + $b ^ 3$ + (-1) $^ 3$ - 3 (a)(b)(-1) = 0

Now a + b - 1 = 0 means a + b = 1

or $a ^ 2$ + $b ^ 2$ + (-1) $^2$ - ab - b(-1) - a(-1) = 0 means a = b = -1 means a + b = -2

Therefore x = -2 and y = 1 or 3x + 30y = 24