To factorise, the first step is to make the expression equal to zero, then try to solve it. So if we put 8x^3-y^6=0, we can write 8x^3=y^6. Taking cube roots of each side we get 2x=y^2. That tells us that 2x-y^2 is a factor. To find the other factor(s), we use algebraic long division to divide (2x-y^2) into 8x^3-y^6. We look at 2x in the divisor and ask: how many times does it divide into 8x^3? It goes 4x^2 times. So we multiply the whole factor by this and we get 8x^3-4x^2y^2. Now subtract this from the dividend to get 4x^2y^2-y^6. The first term in the quotient is the multiplier 4x^2. Now we ask: how many times does (2x...) go into the remainder 4x^2y^2...? It's 2xy^2, so multiply the divisor by this to get 4x^2y^2-2xy^4. Subtract from the remainder and we get 2xy^4-y^6. The quotient in the meantime gains another term, the last multiplier, and becomes 4x^2+2xy^2. Finally, divide the divisor into the new remainder and we can see it goes exactly y^4 times, so the quotient has become 4x^2+2xy^2+y^4. This doesn't factorise further so the answer is: (2x-y^2)(4x^2+2xy^2+y^4)=8x^3-y^6.