NCERT Solutions for Class 9 Maths Exercise 2.4 Question 10
Understanding the Question 🧐
This question is a direct application of the “sum of cubes” and “difference of cubes” identities that we verified in Question 9. We need to factorise expressions that consist of two terms, each of which is a perfect cube. These ncert solutions will guide you through applying the correct formulas.
Factorise each of the following:
(i) &&27y^3 + 125z^3&&
(ii) &&64m^3 – 343n^3&&
Hint: Remember the identities from the previous question:
- &&x^3 + y^3 = (x+y)(x^2 – xy + y^2)&&
- &&x^3 – y^3 = (x-y)(x^2 + xy + y^2)&&
Part (i): Factorise &&27y^3 + 125z^3&& 📝
Step 1: Identify the correct identity.
The expression is a sum of two terms, so we will use the sum of cubes identity: &&x^3 + y^3 = (x+y)(x^2 – xy + y^2)&&.
Step 2: Rewrite each term as a perfect cube.
We need to find what is being cubed in each term.
&&27y^3 = 3 \times 3 \times 3 \times y \times y \times y = (3y)^3&&
&&125z^3 = 5 \times 5 \times 5 \times z \times z \times z = (5z)^3&&
Step 3: Determine the values of &&x&& and &&y&&.
By comparing &&(3y)^3 + (5z)^3&& with &&x^3 + y^3&&, we get:
&&x = 3y&& and &&y = 5z&&.
Step 4: Substitute &&x&& and &&y&& into the identity.
Now, we plug these values into the formula &&(x+y)(x^2 – xy + y^2)&&.
&&= (3y + 5z)((3y)^2 – (3y)(5z) + (5z)^2)&&
Step 5: Simplify the expression.
&&= (3y + 5z)(9y^2 – 15yz + 25z^2)&&
Answer: The factored form of &&27y^3 + 125z^3&& is &&(3y + 5z)(9y^2 – 15yz + 25z^2)&&.
Part (ii): Factorise &&64m^3 – 343n^3&& 📝
Step 1: Identify the correct identity.
This expression is a difference of two terms, so we use the difference of cubes identity: &&x^3 – y^3 = (x-y)(x^2 + xy + y^2)&&.
Step 2: Rewrite each term as a perfect cube.
We need to find the cube roots of &&64&& and &&343&&.
&&64m^3 = 4 \times 4 \times 4 \times m \times m \times m = (4m)^3&&
&&343n^3 = 7 \times 7 \times 7 \times n \times n \times n = (7n)^3&&
Step 3: Determine the values of &&x&& and &&y&&.
Comparing &&(4m)^3 – (7n)^3&& with &&x^3 – y^3&&, we have:
&&x = 4m&& and &&y = 7n&&.
Step 4: Substitute &&x&& and &&y&& into the identity.
We plug these values into the formula &&(x-y)(x^2 + xy + y^2)&&.
&&= (4m – 7n)((4m)^2 + (4m)(7n) + (7n)^2)&&
Step 5: Simplify the expression.
&&= (4m – 7n)(16m^2 + 28mn + 49n^2)&&
Answer: The factored form of &&64m^3 – 343n^3&& is &&(4m – 7n)(16m^2 + 28mn + 49n^2)&&.
- Same: The sign in the first bracket is the same as in the problem.
- Opposite: The sign of the first term in the second bracket is opposite to the sign in the problem.
- Always Positive: The last sign is always positive.
- The first step is always to express the given terms as perfect cubes.
- Memorizing the cubes of numbers from 1 to 10 (e.g., &&4^3=64&&, &&5^3=125&&, &&7^3=343&&) will make these problems much faster.
- Carefully substitute the values of &&x&& and &&y&& into the correct identity.
FAQ
Q: Which algebraic identities are used to solve this question?
A: This question uses the ‘sum of cubes’ and ‘difference of cubes’ identities:
1. &&x^3 + y^3 = (x+y)(x^2 – xy + y^2)&&
2. &&x^3 – y^3 = (x-y)(x^2 + xy + y^2)&&.
These were verified in the previous question (Question 9).
Q: How do you identify which formula to use for a given problem?
A: You can identify the correct formula by the sign between the two cubic terms. If there is a plus sign (&&+&&), use the sum of cubes formula. If there is a minus sign (&&-&&), use the difference of cubes formula.
Q: What is the first step to factorise an expression like &&27y^3 + 125z^3&&?
A: The first step is to rewrite each term as a perfect cube. For example, &&27y^3&& should be written as &&(3y)^3&& and &&125z^3&& should be written as &&(5z)^3&&. This helps you identify the ‘&&x&&’ and ‘&&y&&’ values for the formula.
Q: What is the cube root of 343?
A: The cube root of &&343&& is &&7&&, because &&7 \times 7 \times 7 = 343&&.
Q: How does the SOAP mnemonic help in these factorisations?
A: The SOAP mnemonic (Same, Opposite, Always Positive) helps you remember the signs in the factored form. For &&x^3 + y^3&&, the signs are &&(x+y)(x^2 – xy + y^2)&&, which follow the Same (&&+&&), Opposite (&&-&&), Always Positive (&&+&&) pattern.
Q: What is the complete factored form of &&64m^3 – 343n^3&&?
A: The factored form is &&(4m – 7n)(16m^2 + 28mn + 49n^2)&&.
Further Reading
For more examples on polynomial factorisation, consult the official NCERT Class 9 textbook on their website: https://ncert.nic.in/
