NCERT Solutions for Class 9 Maths Exercise 2.4 Question 11
Understanding the Question 🧐
This problem requires us to factorise a specific type of polynomial. The expression has three cubic terms and one term that is a product of the variables. This structure is a strong hint to use a special algebraic identity. These ncert solutions will break down the process of applying this identity.
Factorise: &&27x^3 + y^3 + z^3 – 9xyz&&
The key to solving this is recognizing that the expression fits the form of the identity:
&&a^3 + b^3 + c^3 – 3abc = (a+b+c)(a^2 + b^2 + c^2 – ab – bc – ca)&&
Factorising &&27x^3 + y^3 + z^3 – 9xyz&& 📝
Step 1: Rewrite the given expression in the form &&a^3 + b^3 + c^3 – 3abc&&.
We need to identify the base of each cubic term and check if the last term matches the &&-3abc&& format.
- &&27x^3 = (3x)^3&&
- &&y^3 = (y)^3&&
- &&z^3 = (z)^3&&
So, our potential &&a, b,&& and &&c&& are &&3x, y,&& and &&z&& respectively. Now let’s check the last term:
&&-9xyz = -3 \cdot (3x) \cdot (y) \cdot (z)&&
Since it matches the &&-3abc&& format, we can confirm our choices.
Step 2: Determine the values of &&a, b,&& and &&c&&.
By comparing &&(3x)^3 + (y)^3 + (z)^3 – 3(3x)(y)(z)&& with the identity, we have:
&&a = 3x&&, &&b = y&&, &&c = z&&
Step 3: Substitute these values into the RHS of the identity.
The identity’s factored form (RHS) is &&(a+b+c)(a^2 + b^2 + c^2 – ab – bc – ca)&&. Let’s substitute our values into this.
&&= (3x + y + z)((3x)^2 + (y)^2 + (z)^2 – (3x)(y) – (y)(z) – (z)(3x))&&
Step 4: Simplify the resulting expression.
Now, we just need to perform the squaring and multiplication inside the second bracket.
&&= (3x + y + z)(9x^2 + y^2 + z^2 – 3xy – yz – 3zx)&&
Answer: The factored form of &&27x^3 + y^3 + z^3 – 9xyz&& is &&(3x + y + z)(9x^2 + y^2 + z^2 – 3xy – yz – 3zx)&&.
- Memorize this important identity: &&a^3 + b^3 + c^3 – 3abc = (a+b+c)(a^2 + b^2 + c^2 – ab – bc – ca)&&.
- The first step is always to rewrite the given expression to explicitly show the cubed terms, like writing &&27x^3&& as &&(3x)^3&&.
- Be careful with the signs in the second bracket of the factored form: the product terms (&&ab, bc, ca&&) are all negative.
FAQ
Q: What is the key algebraic identity used to solve this question?
A: The key identity used for this factorisation is:
&&a^3 + b^3 + c^3 – 3abc = (a+b+c)(a^2 + b^2 + c^2 – ab – bc – ca)&&.
Q: How do you recognize that this specific identity should be used?
A: You can recognize the pattern by observing that the expression contains three terms that are perfect cubes (&&27x^3, y^3, z^3&&) and a fourth term (&&-9xyz&&) which is &&-3&& times the product of the cube roots of the first three terms.
Q: What is the first step in factorising &&27x^3 + y^3 + z^3 – 9xyz&&?
A: The first step is to rewrite each term of the expression to match the form of the identity &&a^3 + b^3 + c^3 – 3abc&&. This means writing &&27x^3&& as &&(3x)^3&&, &&y^3&& as &&(y)^3&&, &&z^3&& as &&(z)^3&&, and &&9xyz&& as &&3(3x)(y)(z)&&.
Q: In the expression &&27x^3 + y^3 + z^3 – 9xyz&&, what are the values of ‘a’, ‘b’, and ‘c’ for the identity?
A: After rewriting the expression, we can see that &&a = 3x&&, &&b = y&&, and &&c = z&&.
Q: What is the final factored form of the expression?
A: The final factored form is &&(3x+y+z)(9x^2 + y^2 + z^2 – 3xy – yz – 3zx)&&.
Q: Is there a special case for this identity?
A: Yes, a very important special case is when &&(a+b+c) = 0&&. In this situation, the entire right side of the identity becomes zero, which implies that &&a^3 + b^3 + c^3 = 3abc&&.
Further Reading
For more details on polynomial identities, refer to the official NCERT Class 9 Maths textbook on their website: https://ncert.nic.in/
