NCERT Solutions for Class 9 Maths Exercise 2.4 Question 6
Understanding the Question 🧐
This question asks us to expand several cubic expressions. These are binomials raised to the power of &&3&&. To solve this, we will use two fundamental algebraic identities. This guide provides detailed ncert solutions to help you master these expansions.
The key identities we’ll use are:
1. For sums: &&(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3&&
2. For differences: &&(a – b)^3 = a^3 – 3a^2b + 3ab^2 – b^3&&
Write the following cubes in expanded form:
Part (i): Expand &&(2x + 1)^3&& 📝
Step 1: Identify the correct identity.
The expression is of the form &&(a + b)^3&&. So, we will use the identity:
&&(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3&&
Step 2: Identify the terms ‘a’ and ‘b’.
Comparing &&(2x + 1)^3&& with &&(a + b)^3&&, we get:
&&a = 2x&&
&&b = 1&&
Step 3: Substitute the values into the identity and simplify.
Substitute &&a = 2x&& and &&b = 1&& into the formula:
&& (2x + 1)^3 = (2x)^3 + 3(2x)^2(1) + 3(2x)(1)^2 + (1)^3 &&
Now, let’s compute each term carefully:
&& (2x)^3 = 2^3 \cdot x^3 = 8x^3 &&
&& 3(2x)^2(1) = 3(4x^2)(1) = 12x^2 &&
&& 3(2x)(1)^2 = 3(2x)(1) = 6x &&
&& (1)^3 = 1 &&
Step 4: Combine the terms to get the final answer.
Putting it all together, we get:
&&(2x + 1)^3 = 8x^3 + 12x^2 + 6x + 1&&
Part (ii): Expand &&(2a – 3b)^3&& 📝
Step 1: Identify the correct identity.
The expression is of the form &&(x – y)^3&&. The identity is:
&&(x – y)^3 = x^3 – 3x^2y + 3xy^2 – y^3&&
Step 2: Identify the terms ‘x’ and ‘y’.
Comparing &&(2a – 3b)^3&& with &&(x – y)^3&&, we have:
&&x = 2a&&
&&y = 3b&&
Note: ‘y’ is &&3b&&, not &&-3b&&. The negative sign is already in the formula.
Step 3: Substitute and simplify.
Substitute &&x = 2a&& and &&y = 3b&& into the formula:
&& (2a – 3b)^3 = (2a)^3 – 3(2a)^2(3b) + 3(2a)(3b)^2 – (3b)^3 &&
Let’s calculate each part:
&& (2a)^3 = 8a^3 &&
&& 3(2a)^2(3b) = 3(4a^2)(3b) = 36a^2b &&
&& 3(2a)(3b)^2 = 3(2a)(9b^2) = 54ab^2 &&
&& (3b)^3 = 27b^3 &&
Step 4: Combine the terms.
The final expanded form is:
&&(2a – 3b)^3 = 8a^3 – 36a^2b + 54ab^2 – 27b^3&&
Part (iii): Expand &&\left(\frac{3}{2}x + 1\right)^3&& 📝
Step 1: Identify the identity.
This is of the form &&(a + b)^3&&. The identity is:
&&(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3&&
Step 2: Identify ‘a’ and ‘b’.
Here, we have:
&&a = \frac{3}{2}x&&
&&b = 1&&
Step 3: Substitute and simplify.
Substitute these values:
&& \left(\frac{3}{2}x + 1\right)^3 = \left(\frac{3}{2}x\right)^3 + 3\left(\frac{3}{2}x\right)^2(1) + 3\left(\frac{3}{2}x\right)(1)^2 + (1)^3 &&
Now, compute each term:
&& \left(\frac{3}{2}x\right)^3 = \frac{3^3}{2^3}x^3 = \frac{27}{8}x^3 &&
&& 3\left(\frac{3}{2}x\right)^2(1) = 3\left(\frac{9}{4}x^2\right)(1) = \frac{27}{4}x^2 &&
&& 3\left(\frac{3}{2}x\right)(1)^2 = \frac{9}{2}x &&
&& (1)^3 = 1 &&
Step 4: Combine to get the final answer.
The expanded form is:
&&\left(\frac{3}{2}x + 1\right)^3 = \frac{27}{8}x^3 + \frac{27}{4}x^2 + \frac{9}{2}x + 1&&
Part (iv): Expand &&\left(x – \frac{2}{3}y\right)^3&& 📝
Step 1: Identify the identity.
This expression is in the form &&(a – b)^3&&. The identity is:
&&(a – b)^3 = a^3 – 3a^2b + 3ab^2 – b^3&&
Step 2: Identify ‘a’ and ‘b’.
Comparing &&\left(x – \frac{2}{3}y\right)^3&& with &&(a – b)^3&&, we get:
&&a = x&&
&&b = \frac{2}{3}y&&
Step 3: Substitute and simplify.
Substitute the values into the formula:
&& \left(x – \frac{2}{3}y\right)^3 = (x)^3 – 3(x)^2\left(\frac{2}{3}y\right) + 3(x)\left(\frac{2}{3}y\right)^2 – \left(\frac{2}{3}y\right)^3 &&
Let’s calculate each term:
&& (x)^3 = x^3 &&
&& 3(x)^2\left(\frac{2}{3}y\right) = \frac{6}{3}x^2y = 2x^2y &&
&& 3(x)\left(\frac{2}{3}y\right)^2 = 3(x)\left(\frac{4}{9}y^2\right) = \frac{12}{9}xy^2 = \frac{4}{3}xy^2 &&
&& \left(\frac{2}{3}y\right)^3 = \frac{8}{27}y^3 &&
Step 4: Combine the terms.
The final expansion is:
&&\left(x – \frac{2}{3}y\right)^3 = x^3 – 2x^2y + \frac{4}{3}xy^2 – \frac{8}{27}y^3&&
Conclusion and Key Points ✅
In this question, we successfully expanded four cubic binomials using the standard algebraic identities. The key was to correctly identify which identity to use (sum or difference) and to carefully substitute the terms for &&a&& and &&b&& (or &&x&& and &&y&&).
- Identity 1 (Sum): &&(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3&&
- Identity 2 (Difference): &&(a – b)^3 = a^3 – 3a^2b + 3ab^2 – b^3&&
- When cubing a term like &&(kx)^3&&, remember to cube both the coefficient &&k&& and the variable &&x&&. For example, &&(2x)^3 = 8x^3&&.
- Be extra careful with signs and fractions during substitution and simplification to avoid common errors.
FAQ
Q: What is the formula for expanding &&(a + b)^3&&?
A: The algebraic identity for expanding &&(a + b)^3&& is &&a^3 + 3a^2b + 3ab^2 + b^3&&. This formula is essential for finding the cube of any binomial sum.
Q: How do you expand &&(2x + 1)^3&& using the identity?
A: To expand &&(2x + 1)^3&&, you use the identity &&(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3&&. Here, you set &&a = 2x&& and &&b = 1&&. Substituting these values into the formula gives the final expanded form: &&8x^3 + 12x^2 + 6x + 1&&.
Q: What is the identity for &&(a – b)^3&&?
A: The identity for &&(a – b)^3&& is &&a^3 – 3a^2b + 3ab^2 – b^3&&. This is used to expand the cube of a binomial where one term is subtracted from another.
Q: How do you handle fractions in cubic expansions, like in &&\left(\frac{3}{2}x + 1\right)^3&&?
A: When expanding an expression with fractions like &&\left(\frac{3}{2}x + 1\right)^3&&, you treat the entire fractional term as one variable. In this case, &&a = \frac{3}{2}x&& and &&b = 1&&. When you cube the fractional term, you must cube both the numerator and the denominator. For instance, &&\left(\frac{3}{2}x\right)^3 = \frac{3^3}{2^3}x^3 = \frac{27}{8}x^3&&.
Q: What is a common mistake when expanding an expression like &&(2a – 3b)^3&&?
A: A very common mistake is confusing the signs. When using the identity &&(x – y)^3 = x^3 – 3x^2y + 3xy^2 – y^3&&, you should identify &&x = 2a&& and &&y = 3b&&. The negative sign is already accounted for in the formula’s structure. Substituting &&y = -3b&& is incorrect and will lead to errors in the signs of the final terms.
Further Reading
For more information on polynomials and algebraic identities, you can refer to the official NCERT textbook. A digital version is available at the NCERT website: https://ncert.nic.in/.
