n in Figure 5.1(c). The stack in Figure 5.1(c) forms Arrangement 1. The total volume of this arrangement of cubes is calculated (see the calculations).
Math Labs with Activity - Verify the Identity a³ - b³ =(a-b)(a² + ab+b²) 1
Step 2: We shall make Arrangement 2 for 26 cubes.
This arrangement consists of three stacks—the first stack [shown in Figure 5.2(a)] consists of 9 columns of two cubes each, the second stack [shown in Figure 5.2(b)] consists of two rows of three cubes each, and the third stack consists of 1 row of 2 cubes.
The total volume of this arrangement of cubes is calculated (see the calculations).
Math Labs with Activity - Verify the Identity a³ - b³ =(a-b)(a² + ab+b²) 2
Observations
Since the two arrangements have equal number of cubes (each arrangement has 26 cubes) and all the cubes have the same volume (1 cubic unit), the total volumes in both the arrangements must be equal.
Calculations
Volume of Arrangement 1
Volume of the stack in Figure 5.1(a) = a³.
Volume of the stack in Figure 5.1(b) =b³.
=volume of Arrangement 1 volume of the stack in Figure 5.1(c)
= volume of the stack in Figure 5.1(a)
=volume of the stack in Figure 5.1(b) = a³-b³.
Volume of Arrangement 2
Volume of the stack in Figure 5.2(a)=(a-b)a².
Volume of the stack in Figure 5.2(b) =(a-b)ab.
Volume of the stack in Figure 5.2(c) = (a-b)b².
volume of Arrangement 2 = (a-b)a² + (a -b)ab + (a -b)b².
=(a-b)(a² + ab+b²).
Since the total volume in the two arrangements must be the same, therefore
a³ – b³ =(a-b)(a² + ab+b²).
Result
It is verified that a³ – b³ =(a-b)(a² + ab+b²).
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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Answers & Comments
Answer:
a³+b³= (a+b)(a²-ab+b²)
Step-by-step explanation:
verify a³+b³= (a+b)(a²-ab+b²)
RHS= (a + b) (a² - ab + b²)
= a(a² - ab + b²) + b(a² - ab + b²)
= a(a²) - a(ab) + a(b²) + b(a²) - b (ab) + b(b²)
= a³ - a²b + ab² +a²b - ab² + b³
Rearrange similar terms
= a³ - a²b +a²b + ab² - ab² + b³
= a³+b³
= LHS
a³+b³= (a+b)(a²-ab+b²)
Hence verified.
Answer:
n in Figure 5.1(c). The stack in Figure 5.1(c) forms Arrangement 1. The total volume of this arrangement of cubes is calculated (see the calculations).
Math Labs with Activity - Verify the Identity a³ - b³ =(a-b)(a² + ab+b²) 1
Step 2: We shall make Arrangement 2 for 26 cubes.
This arrangement consists of three stacks—the first stack [shown in Figure 5.2(a)] consists of 9 columns of two cubes each, the second stack [shown in Figure 5.2(b)] consists of two rows of three cubes each, and the third stack consists of 1 row of 2 cubes.
The total volume of this arrangement of cubes is calculated (see the calculations).
Math Labs with Activity - Verify the Identity a³ - b³ =(a-b)(a² + ab+b²) 2
Observations
Since the two arrangements have equal number of cubes (each arrangement has 26 cubes) and all the cubes have the same volume (1 cubic unit), the total volumes in both the arrangements must be equal.
Calculations
Volume of Arrangement 1
Volume of the stack in Figure 5.1(a) = a³.
Volume of the stack in Figure 5.1(b) =b³.
=volume of Arrangement 1 volume of the stack in Figure 5.1(c)
= volume of the stack in Figure 5.1(a)
=volume of the stack in Figure 5.1(b) = a³-b³.
Volume of Arrangement 2
Volume of the stack in Figure 5.2(a)=(a-b)a².
Volume of the stack in Figure 5.2(b) =(a-b)ab.
Volume of the stack in Figure 5.2(c) = (a-b)b².
volume of Arrangement 2 = (a-b)a² + (a -b)ab + (a -b)b².
=(a-b)(a² + ab+b²).
Since the total volume in the two arrangements must be the same, therefore
a³ – b³ =(a-b)(a² + ab+b²).
Result
It is verified that a³ – b³ =(a-b)(a² + ab+b²).
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.
Math Labs with ActivityMath LabsMath Lab ManualScience LabsScience Practical Skills
Filed Under: CBSE
Primary Sidebar
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RS Aggarwal Solutions
RS Aggarwal Solutions Class 10
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Concise Mathematics Class 10 ICSE Solutions
Concise Physics Class 10 ICSE Solutions
Concise Chemistry Class 10 ICSE Solutions
Concise Biology Class 10 ICSE Solutions
Concise Mathematics Class 9 ICSE Solutions
Concise Physics Class 9 ICSE Solutions
Concise Chemistry Class 9 ICSE Solutions
Concise Biology Class 9 ICSE Solutions
ML Aggarwal Solutions
ML Aggarwal Class 10 Solutions
ML Aggarwal Class 9 Solutions
ML Aggarwal Class 8 Solutions
ML Aggarwal Class 7 Solutions
ML Aggarwal Class 6 Solutions
HSSLive Plus One
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Kerala SSLC
RECENT POSTS
Wireless Network Advantages and Disadvantages | Advantages and Disadvantages of Wireless Network Communication
5 Modes of Transportation and Their Advantages and Disadvantages
GST Advantages and Disadvantages | Merits and Demerits, Pros and Cons, Benefits and Drawbacks
Genetically Modified Crops Advantages and Disadvantages | Advantages and Disadvantages of GM Crops
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Advantages and Disadvantages of Internet | Top 5 Advantages and Disadvantages of Internet Communication
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