I can solve any math equation.

           ____
x = ±√3y+1
Where the range of y is:
y ≥ -1/3

           ____
x = ±√3y+1

Auming that:
a ≥ 1, b ≥ 1, c ≥ 1

Let n be 3 first...
Let a_k be a sequence of numbers where a_k = k³

Finding the difference of each term, you should get
a_k+1-a_k =
(k-1)³-k³ =
3k²+3k+1

Now find a cube that equals to 3k²+3k+1.
The only solution that can be found is 

But that solution isn't even an integer.

Therefore, there is no such combination when a ≥ 1, b ≥ 1, c ≥ 1

thx!

This is an equation of a circle.
The standard equation of a circle is (x-x₀)²+(y-y₀)²=r²
Rearranging the equation to make it that way..
(x+9/2)²+(y+1)² = 85/4                                                                   __
The center of the circle is (-9/2,-1), and the radius of the circle is √85/2

Assuming the n is an integer...