Math Labs with Activity – Verify the Identity (a-b)³ =a³ – 3a²b + 3ab² -b³

Math Labs with Activity – Verify the Identity (a-b)³ =a³ – 3a²b + 3ab² -b³

OBJECTIVE

To verify the identity (a-b)³ =a³ – 3a²b + 3ab² -b³ using a set of unit cubes.

Materials Required
A set of 128 plastic cubes each having dimensions (1 unit x 1 unit x 1 unit).

Procedure
We shall verify the identity for a = 4 and b = 1.
Step 1: We shall make Arrangement 1 for 64 cubes.
This arrangement consists of a stack formed by 16 columns of 4 cubes each, as shown in Figure 7.1. The total volume of this arrangement of cubes is calculated (see the calculations).

Step 2: We shall make Arrangement 2 for 64 cubes.
This arrangement consists of 8 stacks. Stack I shown in Figure 7.2(a) consists of 9 columns of 3 cubes each. Stacks II, III and IV shown in Figure 7.2(b) consist of 3 columns of 3 cubes each.
Stacks V, VI and VII shown in Figure 7.2(c) consist of 1 column of 3 cubes each. Stack VIII shown in Figure 7.2(d) consists of only one cube.The volume of each stack is calculated and these volumes are added up to get the total volume of this arrangement of cubes (see the calculations).

Observations
Since the two arrangements have equal number of cubes (each arrangement has 64 cubes) and all the cubes have the same volume (1 cubic unit), the total volume in both the arrangements must be the same.

Calculations

1. Volume of Arrangement 1
   Volume of the stack in Figure 7.1= a³.
   volume of Arrangement 1 of the cubes = a³.
2. Volume of Arrangement 2
   Volume of Stack I [the stack in Figure 7.2(a)] = (a -b)³.
   Volume of each of Stacks II, IE and IV=(a -b)²b.
   total volume of the three stacks [shown in Figure 7.2(b)] = 3(a-b)²b. Volume of each of Stacks V, VI and VII = (a-b)b².
   total volume of the three stacks [shown in Figure 7.2(c)] = 3(a -b)b². Volume of stack VIII = b³.
   Thus, total volume of Arrangement 2 of the cubes
   =(a-b)³ + 3(a-b)²b+3(a-b)b²+b³.
   Since the total volume in the two arrangements must be the same, therefore
   a³=(a-b)³ + 3(a-b)²b + 3(a-b)b²+b³ => (a-b)³=a³-3(a-b)²b-3(a-b)b²-b³
   => (a-b)³ = a³-3(a-b)((a-b)b+b²)-b³
   => (a-b)³ = a³-3ab(a-b)-b³
   => (a-b)³ =a³ – 3a²b + 3ab² -b³.

Result
It is verified that (a-b)³ =a³ – 3a²b + 3ab² -b³.

Remarks: The students must try to verify this identity for other values of a and b by taking required number of cubes and arranging them suitably.

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