NCERT Solutions for Class 9 Maths Exercise 2.4 Question 7
Understanding the Question 🧐
In this problem, we need to find the value of some numbers raised to the power of &&3&&. Instead of doing lengthy multiplication, we are asked to use suitable algebraic identities to make the calculation easier. This is a common technique in algebra that simplifies complex arithmetic. These ncert solutions will guide you through each step.
Evaluate the following using suitable identities:
(i) &&(99)^3&&
(ii) &&(102)^3&&
(iii) &&(998)^3&&
The key here is to rewrite each number as a sum or difference of two simple numbers (usually one of them being a power of 10 like 100 or 1000) and then apply the cubic identities.
Part (i): Evaluate &&(99)^3&& 📝
Step 1: Rewrite the expression
We can write &&99&& as &&(100 – 1)&&. This is much easier to work with. So, &&(99)^3 = (100 – 1)^3&&.
Step 2: Identify and apply the suitable identity
The expression &&(100 – 1)^3&& is in the form of &&(a – b)^3&&.
The identity is: &&(a – b)^3 = a^3 – b^3 – 3ab(a – b)&&.
Here, we have &&a = 100&& and &&b = 1&&.
Step 3: Substitute the values and calculate
Substitute &&a = 100&& and &&b = 1&& into the identity:
&&(100 – 1)^3 = (100)^3 – (1)^3 – 3(100)(1)(100 – 1)&&
Now, let’s calculate each part:
&&= 1000000 – 1 – 300(99)&&
&&= 1000000 – 1 – 29700&&
Be careful with the subtraction order. First, combine the negative terms.
&&= 1000000 – 29701&&
&&= 970299&&
Answer: Thus, &&(99)^3 = 970299&&.
Part (ii): Evaluate &&(102)^3&& 📝
Step 1: Rewrite the expression
We can express &&102&& as &&(100 + 2)&&. So, &&(102)^3 = (100 + 2)^3&&.
Step 2: Identify and apply the suitable identity
The expression &&(100 + 2)^3&& matches the form of &&(a + b)^3&&.
The identity is: &&(a + b)^3 = a^3 + b^3 + 3ab(a + b)&&.
In this case, &&a = 100&& and &&b = 2&&.
Step 3: Substitute the values and calculate
Putting &&a = 100&& and &&b = 2&& into the formula:
&&(100 + 2)^3 = (100)^3 + (2)^3 + 3(100)(2)(100 + 2)&&
Calculate the terms:
&&= 1000000 + 8 + 600(102)&&
&&= 1000000 + 8 + 61200&&
&&= 1061208&&
Answer: Therefore, &&(102)^3 = 1061208&&.
Part (iii): Evaluate &&(998)^3&& 📝
Step 1: Rewrite the expression
We can write &&998&& as &&(1000 – 2)&&. This gives us &&(998)^3 = (1000 – 2)^3&&.
Step 2: Identify and apply the suitable identity
This expression is in the form of &&(a – b)^3&&.
We use the identity: &&(a – b)^3 = a^3 – b^3 – 3ab(a – b)&&.
Here, &&a = 1000&& and &&b = 2&&.
Step 3: Substitute the values and calculate
Substitute &&a = 1000&& and &&b = 2&&:
&&(1000 – 2)^3 = (1000)^3 – (2)^3 – 3(1000)(2)(1000 – 2)&&
Now, let’s compute the values:
&&= 1000000000 – 8 – 6000(998)&&
&&= 1000000000 – 8 – 5988000&&
&&= 1000000000 – 5988008&&
&&= 994011992&&
Answer: Hence, &&(998)^3 = 994011992&&.
Conclusion and Key Points ✅
By using the algebraic identities for the cube of a binomial, we can easily evaluate powers of numbers that are close to a base of &&10&&. This method is efficient and reduces the chances of calculation errors. The two key identities used were:
- &&(a + b)^3 = a^3 + b^3 + 3ab(a + b)&&
- &&(a – b)^3 = a^3 – b^3 – 3ab(a – b)&&
- Identify whether you need the &&(a+b)^3&& or &&(a-b)^3&& identity based on how you rewrite the number.
- Pay close attention to the signs. The &&(a-b)^3&& formula has negative terms, which are a common source of mistakes.
- Double-check your calculations, especially when dealing with large numbers and multiple zeros.
FAQ
Q: What is the goal of evaluating expressions like &&(99)^3&& using identities?
A: The goal is to calculate the value of cubic expressions without performing tedious direct multiplication. By using algebraic identities, we can simplify the calculation, making it faster and less prone to errors.
Q: Which identity is used to evaluate &&(99)^3&&?
A: To evaluate &&(99)^3&&, we first rewrite &&99&& as &&(100 – 1)&&. Then, we apply the algebraic identity &&(a-b)^3 = a^3 – b^3 – 3ab(a-b)&&, with &&a=100&& and &&b=1&&.
Q: Which identity is used for &&(102)^3&&?
A: For &&(102)^3&&, we rewrite &&102&& as &&(100 + 2)&& and use the identity &&(a+b)^3 = a^3 + b^3 + 3ab(a+b)&&, setting &&a=100&& and &&b=2&&.
Q: How do you choose the base for rewriting the numbers?
A: Choose a convenient base that is a power of &&10&& (like &&100&& or &&1000&&) and is closest to the given number. This makes the second term in the binomial small, simplifying the calculations significantly.
Q: Why not just multiply &&99 \times 99 \times 99&& directly?
A: While direct multiplication is possible, the question specifically requires using ‘suitable identities’. This method is a key algebraic skill that simplifies complex calculations and is often faster and more accurate for mental math.
Q: What is a common mistake when using the &&(a-b)^3&& identity?
A: A common mistake is messing up the signs. Remember that in the expansion of &&(a-b)^3&&, the &&b^3&& term and the &&3ab(a-b)&& term are both subtracted from &&a^3&&.
Further Reading
For more information on polynomials and algebraic identities, you can refer to the official NCERT textbook and resources available on their website: https://ncert.nic.in/
