The main thing to notice on this problem is that we can write $\left({2y-6}\right)^3$ as $$\left({2y-6}\right)^3 = \left({2y-6}\right)\left({2y-6}\right)^2$$ This means that we can use the FOIL technique on the second binomial multiplication, and then multiply the first factor ${2y-6}$ by the result. $$\begin{aligned} \left({2y-6}\right)^3 & = \left({2y-6}\right)\left[\left({2y-6}\right)\left({2y-6}\right)\right]\\ &=\left({2y-6}\right)\left[{4y^{2}-12y-12y+36}\right]\\ &=\left({2y-6}\right)\left[{4y^{2}-24y+36}\right]\\ & = {8y^{3}-48y^{2}+72y-24y^{2}+144y-216}\\ &={8y^{3}-72y^{2}+216y-216} \end{aligned}$$ You might like to know that the formula for the cube of the difference of two terms is $$(a-b)^3 = a^3 - 3a^2b+3ab^2-b^3$$ If you have the time, you can verify that this formula works in this problem as an exercise.