Surface Area and Volume relationship

Question

I know that the $SA = 6s^2$ and that the volume is equal to the base $x$ the $side = s^3$. However, I'm not sure how to approach this though.

Answer 1

Let $a$ be the sidelength of the first cube. Let $b$ be the sidelength of the second cube.

The surface area for a cube with sidelength $s$ is given by the formula $\text{SurfaceArea}=6s^2$ (note it is quadratic in $s$, not cubic as in your typo above)

The volume of a cube with sidelength $s$ is given by the formula $\text{vol}(S) = s^3$

We are told that $\text{vol}(A)=27\cdot\text{vol}(B)$ (worded above as "the volume of A is 27 times that of the volume of B")

$$\text{vol}(A)=27\cdot\text{vol}(B)\\ a^3 = 27b^3 \\ \sqrt[3]{a^3} = \sqrt[3]{27b^3}\\ a = 3b$$

We are curious as to the ratio between the surface areas:

$\text{SurfaceArea}(A) = 6a^2 = 6(3b)^2 = 6\cdot 9\cdot b^2 = 9\cdot (6b^2) = 9\cdot\text{SurfaceArea}(B)$

Thus, the ratio is $9:1$ (since surface(A)=9surface(B), A's surface area is larger)

Answer 2

$$\frac{r_1^3}{r_2^3}=27$$ $$r_1=3r_2$$ $$\frac{A_1}{A_2}=\frac{6(3r_2)^2}{6r_2^2}=9$$