Question

The edge of one cube is 4 m shorter than the edge of a second cube. The volumes of the two cubes differ by 1216 m^3. Find the edge of the smaller cube.

Answer 1

Volume of cube, V = edge^3
Let edge of cube#1 = (x-4) m, therefore volume of cube#1, v1 = (x-4)^3 m
Let edge of cube#2 = x m, therefore volume of cube#2, v2 = x^3 m
Diff. in volume (in m) = 1216 = v2-v1 = [ x^3 - (x-4)^3 ] 
= x^3 - [(x-4)(x-4)(x-4)]
= x^3  - [x^2 - 8x +16(x - 4)]
=  x^3 - [ x^3 - 12x^2 + 48x - 64 ]
= 12x^2 - 48x + 64
= 4 (3x^2 - 12x + 16)
Therefore 4 (3^2 - 12x + 16) = 1216
3x^2 - 12x + 16 = 1216/4 = 304
3x^2 - 12x - 288 = 0
3 (x^2 - 4x - 96) = 0
(x^2 - 4x - 96) = 0
(x - 12) (x + 8) =0
(x-12) = 0
Therefore x = 12 m 
Edge of cube#2 = x m = 12m
Edge of cube#1 = (x-4) m = 8m

Answer 2

$a- the\ edge\ of\ the\ smaller\ cube\ \ \ \wedge\ \ \ a>0\\V_a-the\ volume\ of\ the\ smaller\ cube\ \ \ \Rightarrow\ \ \ V_a=a^3\\ \\(a+4)^3-a^3=1216\ \ \ \wedge\ \ \ (x+y)^3=x^3+3x^2\cdot y+3x\cdot y^2+y^3\\ \\a^3+3a^2\cdot 4+3a\cdot 4^2+4^3-a^3=1216\\ \\12a^2+48a+64-1216=0\ \ \ \Rightarrow\ \ \ 12a^2+48a-1152=0\ /:12\\ \\a^2-4a-96=0\ \ \ \Rightarrow\ \ \ \Delta=(-4)^2-4\cdot 1\cdot(-96)=16+384=400\\ \\ \sqrt{\Delta} =20\\ \\a_1= \frac{4-20}{2\cdot1} =-80$

$Ans.\ The\ edge\ of\ the\ smaller\ cube=12\ m$