# Calculate the diagonal of a cube

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?

It looks to me as if all your reasoning is correct, but your algebra is off in the final step. $$\frac{6x^2+32}{(\sqrt3x+2)^2}=2 \\ \implies 6x^2+32=2(3x^2+4\sqrt 3 x+4) \\ \implies 32=8\sqrt 3 x+8 \\ \implies 8\sqrt 3 x=24 \\\implies \sqrt 3 x=3$$
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?
• The question explicitly states 'area', not volume, and uses units of cm$^2$, so using $6x^2$ for the total (surface) area is correct. Commented Jan 11, 2020 at 18:32