Calculate the diagonal of a cube

Question

Calculate the diagonal of a cube, knowing that if it is increased by $ 2 $ cm, the total area will increase $ 32 $ $ cm ^ 2 $

Attemp: Let $x$ be the initial sidelength of the cube. Then the diagonal will be $\sqrt3x$ and the total area will be $6x^2$. We see that $\frac{6x^2}{(\sqrt3x)^2}=2$ is a constant value. Then $\frac{6x^2+32}{(\sqrt3x+2)^2}=2 <=> \sqrt3x=1$. The diagonal is $1$.

Correct?

Answer 1

It looks to me as if all your reasoning is correct, but your algebra is off in the final step. $$\frac{6x^2+32}{(\sqrt3x+2)^2}=2 \\ \implies 6x^2+32=2(3x^2+4\sqrt 3 x+4) \\ \implies 32=8\sqrt 3 x+8 \\ \implies 8\sqrt 3 x=24 \\\implies \sqrt 3 x=3 $$

Answer 2

Note that if $d$ is the diagonal of the cube and $V$ the area, and $a=\frac1{3\sqrt3}$, then $V=ad^3$. This is where you went wrong, since you assumed that cubes are $2$-dimensional. So, if $d_0$ is the original diagonal, $d_1$ the new, we know that $$d_0+2=d_1$$ $$ad_0^3+32=ad_1^3$$$$d_0^3+96\sqrt3=(d_0+2)^3=d_0^3+6d_0^2+12d_0+8$$$$6d_0^2+12d_0+(8-96\sqrt3)=0$$Can you solve for $d_0$ from here?