# Cube numbers inquiry

# The prompt

Mathematical inquiry processes: Verify; extend the pattern; generalise and prove. Conceptual field of inquiry: Product of cube numbers; algebraic notation, terms and expressions.

Emmy Bennett, a teacher of mathematics at Priory School, Edgbaston (UK), reports that one of her students designed the prompt:

"Before the half term break my year 8 class were desperate to do an inquiry lesson. Having done some with them last year they really enjoyed the exploration and independence an inquiry lesson gave them. This year one of my pupils, Michael, came up with the prompt we used. This lesson was a mini-inquiry as it only lasted around 30 minutes, but there was still some excellent mathematics being explored. This prompt has the potential to be used to explore algebraic proof with older pupils or simply cube numbers and other powers with younger pupils."

The general case for the two equations can be expressed algebraically to show the difference between the products is n:

### 93 – (8 x 9 x 10) = 729 – 720 = 9

### n3 – [(n – 1)n(n + 1)] = n3 – (n3 – n) = n

Students might change the prompt to explore cases when the numbers in the second equation increase by a difference greater than one:

### n3 – [(n – a)n(n + a)] = n3 – (n3 – a2n) = a2n

December 2020

# Classroom inquiry

The questions and observations from Emmy's year 8 class could lead to multiple lines of inquiry:

Testing different types of number;

Exploring other cubes;

Replacing the consecutive numbers with numbers that have a difference of two and more.

Looking for patterns in the difference between the two products;

Generalising from the pattern;

Expressing the generalisation using algebra and proving;

Changing the first equation to use square numbers.

Myrah and Mariha were exploring lots of examples as well as looking at when he difference between the numbers in the second product was not equal to 1.

Ehsan was exploring negative numbers as well as non-integers. We also discussed algebraic proof.

# Visual representation

The original cube has a volume of (9 cm)3 = 729 cm3. To transform the cube into a cuboid with dimensions 10 cm, 9 cm and 8 cm, a cuboid with dimensions 9 cm, 9 cm and 1 cm (volume 81 cm3) is sliced off the top. Another cuboid with dimensions 1 cm, 9 cm and 8 cm (volume 72 cm3) is attached to the side. So the volume of the new cuboid is 729 – 81 + 72 = 720 cm3.

We can use the same reasoning for any cube of side length (n) that is transformed into a cuboid by increasing the length by 1 cm and reducing the height by 1 cm. To transform the cube (volume n3 cm3) into the cuboid, a cuboid with dimensions n cm, n cm and 1 cm (volume n2 cm3) is sliced off the top. Another cuboid with dimensions 1 cm, n cm and (n – 1) cm [volume (n2 – n) cm3] is attached to the side. So the volume of the new cuboid is n3 – n2 + (n2 – n) = (n3 – n) cm3.

We can use the same reasoning again for any cube of side length (n) that is transformed into a cuboid by increasing the length by a cm and reducing the height by a cm (where n > a). To transform the cube (volume n3 cm3) into the cuboid, a cuboid with dimensions n cm, n cm and a cm (volume an2 cm3) is sliced off the top. Another cuboid with dimensions a cm, n cm and (n – a) cm [volume (an2 – a2n) cm3] is attached to the side. So the volume of the new cuboid is n3 – an2 + (an2 – a2n) = (n3 – a2n) cm3.