cube as funciton of the lenght of diagonal

[barnamah]

I have a question which says:
a)
Find the relationship describing the volume (V) of a cube as a function of the length of the diagonal going through the cube.
(d). You MUST show your work in finding this relationship (by typing it up in Maple).
Enter this as a function in Maple, and evaluate it for a diagonal length of d = 1.2.

[Integr88]

A cube has the volume formula of `V=L*W*H` and because it is cube the all values are the same and L=W=H so `V=L * L * L=L^3`
so volume of cube is `V=L^3`

now the diagonal d of an area is by Pythagorean theorem is `d^2=L^2+W^2` where so the the same way the diagonal of a volume is `d^2=L^2*W^2*H^2` and because we are working on cube and L=W=H so `d^2=L^2+L^2+L^2` and to simplify it `d^2=3L^2` and by taking the square root of both side `sqrt(d^2)=sqrt(3L^2)`so d is `d=sqrt(3L^2)` and because `L^2` is under the square root, it becomes just L and `d=Lsqrt(3)`.

Now we have one formula for volume of cube `V=L^3` and another formula for diagonal `d=Lsqrt(3)` , because we want V as function of diagonal d, then we can write `d=Lsqrt(3)` in terms of L. Dividing both side by `sqrt(3)` we have `d/sqrt(3)=(Lsqrt(3))/sqrt(3)` and both `sqrt(3)` on the right side cancels out and we have
`d/sqrt(3)=L`

If you look at volume we have `V=L^3` so substitute L into this formula and have `V=(d/sqrt(3))^3` and out equation is solve.
`:.` the volume of cube a function of diagonal is `V=(d/sqrt(3))^3`

to evaluate it with d=1.2 (let's assume the unit is meter m) we have
`V=(d/sqrt(3))^3=(1.2/sqrt(3))^3=(1.2/1.732)^3=(0.6928)^3=0.3325` and because the unit is meter so `:.` the volume is `0.3325m^3`

in maple we can define the function as

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f:=d->(d/sqrt(3))^3;

to recall the function f, if we use:

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f(1.2);
we get `0.1920000000sqrt(3)`


to get decimal only use evalf

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evalf(f(1.2));       
0.3325537551