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A cube has edge-length of 8 cm. Cut in into several identical cubes so that their total surface area is 5 times that of the original cube. What is the volume of one such small cube?

I started b finding the surface area of the 8*8*8 cube, which was 384. I multiplied that by 5 to get 1920. I prime factorized 1920 into $2^7 \cdot 3 \cdot 5$. I'm stuck at this point, and it doesn't seem like a cube that can be split up to have these factors.

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    $\begingroup$ Who said the dimensions would be integral? $\endgroup$ – Deepak Nov 20 '16 at 13:43
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    $\begingroup$ What happens to the total area if you divide into $2^3=8$ small cubes? What about $3^3$? What about $4^3$? $\endgroup$ – Arthur Nov 20 '16 at 13:49
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    $\begingroup$ Riffing on Arthur's comment: Imagine the cube is lying on a table. What happens to the total area of the horizontal (those facing up or down) parts of the surfaces, if you cut the cube to $n$ identical layers? Do the same to surfaces facing left/right and front/back. $\endgroup$ – Jyrki Lahtonen Nov 20 '16 at 13:59
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the number of cubes are $$n=(\frac{8}{a})^3$$ so that $a$ is the edge length of small cube

so $$5(6*8*8)=(6a^2)n$$ $$5(6*8*8)=(6a^2)(\frac{8}{a})^3$$ so the $a=\frac{8}{5}$ and $n=125$ the volume of one cube is $(\frac{8}{5})^3$

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Assume $N$ small cubes of side $x$.

You have:

Surface area: $(N)(6)(x^2)= (5)(6)(8^2) \implies Nx^2 = (5)(8^2)$ [equation $1$]

Volume: $Nx^3 = 8^3 \implies (x)(Nx^2) = (8)(8^2)$ [equation $2$]

Divide equation $2$ by equation $1$, what do you have?

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Let y be the length of the edge of the small cube and let the number of small cubes be n. Since total volume of the small cubes is equal to the volume of the larger cube we have 8^3=ny^3. Thus n=(8/y)^3. By question, 5(surface area of large cube)= total surface area of the small cubes Thus 5×6×8×8=(6×y×y×8×8×8)÷(y^3) Solving for y we get y=8÷5=1.6 and n=5^3=125 Also vol of one such small cube is (8/5)^3=512/125

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  • $\begingroup$ Please use Mathjax conventions. First of all, formulas should be enclosed by dollar signs. $\endgroup$ – Jean Marie Nov 23 '16 at 22:04

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