Geometry of Solids Riddle

Question

There are two cubes, each with integer-length sides with this relationship: their combined volumes has the same numerical value as their combined lengths. Find the length of the sides of each cube.

Answer 1

Let Cube A have all 12 sides of length A. Let cube B have all 12 sides of length B.

Vol of A = l*w*h = A * A * A = A³ Vol of B = l*w*h = B * B * B = B³

Combined sides of A = 12 * A = 12A Combined sides of B = 12 * B = 12B

From the problem: "combined volumes" in mathspeak is A³ + B³

"combined [side] lengths" in mathspeak is 12A + 12B

Since they are numerically equivalent ("have the same numeric value") we can set these two bolded expressions equal to each other.

A³ + B³ = 12A + 12B

Factoring each side

( A + B ) (A² - AB + B² ) = 12 ( A + B )

Move everything to one side, then factor out the (A+B) term:

(A+B)(A² - AB + B² - 12) = 0

This means that either A+B=0 -- which cannot be because they both must be positive integers OR

A² - AB + B² - 12 = 0

We can turn this into a quadratic equation by letting A = x, now this becomes x² - Bx + (B²- 12)

This we can use the quadratic formula on:

x = B ± sqrt[ B² - 4(1)(B² - 12) ]

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2(1)

x = B ± sqrt[ B² - 4B² + 48 ]

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2

x = B ± sqrt[ 48 - 3B² ]

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2

Knowing that our cubes' sides cannot be negative nor can they have imaginary lengths, this limits us to solving this quadratic formula with the stuff under the square root CANNOT BE NEGATIVE. So...

48 - 3B² ≥ 0 or 3B² ≤ 48 divide both sides by 3 B² ≤ 16

-4 ≤ B ≤ 4 but we know that the sides of our cubes have to be larger than 0. This limits our inequality to

0 ≤ B ≤ 4 So B can only take on one of the values in the set of {1, 2, 3, or 4}

Trying each of these into the quadratic formula

We find that B=2 works, making x = 4 or A = 4

We find that B = 4 works, making x = 2 or A = 2 so either one of these is the correct answer.

Check. Let A = 4 and B = 2 A³+ B³ = 4³ + 2³ = 64 + 8 = 72

12A + 12B = 12(4) + 12(2) = 48 + 24 = 72

So the correct answer is One cube has sides of length 4 and the other cube has sides of length 2.

Answer 2

a represents the edge of one cube

b represents the edge of the other cube

a³ + b³ = 12a + 12b

a³ + b³ = 12(a + b)

a² - ab + b² = 12

a² - ab + b² - 12 = 0

a = (-b ± √(b² - 4(1)(b² - 12)) / 2

The radicand is -3b² + 12. For integral values it must be a perfect square.

What are the only values of b that make it so?