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Q6(iv):
Write the following cubes in expanded form:
(iv) $(x - \frac{2}{3}y)^3$
Solution :
Initial Setup & Algebraic Identity
We are tasked with expanding the algebraic expression:
$(x - \frac{2}{3}y)^3$
To expand the cube of a binomial difference, we utilize the standard algebraic identity derived from the Binomial Theorem [where $n=3$]. The identity for the cube of a difference is given by:
$(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$
Step 1: Variable Mapping and Substitution
By comparing our given expression $(x - \frac{2}{3}y)^3$ with the standard identity $(a - b)^3$, we can establish the following one-to-one mapping of variables:
- $a = x$
- $b = \frac{2}{3}y$
Substituting these mapped values into the expansion formula yields:
$(x - \frac{2}{3}y)^3 = (x)^3 - 3(x)^2(\frac{2}{3}y) + 3(x)(\frac{2}{3}y)^2 - (\frac{2}{3}y)^3$
Step 2: Term-by-Term Expansion and Simplification
We will now isolate and simplify each of the four terms generated by the expansion [adhering strictly to the order of operations and the laws of exponents, specifically $(xy)^n = x^n y^n$].
-
First Term ($a^3$):
$(x)^3 = x^3$ -
Second Term ($-3a^2b$):
$-3(x)^2(\frac{2}{3}y) = -3 \cdot x^2 \cdot \frac{2}{3}y$
The scalar $3$ in the numerator and the $3$ in the denominator cancel out:
$= -2x^2y$ -
Third Term ($+3ab^2$):
$3(x)(\frac{2}{3}y)^2$
First, square the term inside the parentheses [$(\frac{2}{3})^2 \cdot y^2 = \frac{4}{9}y^2$]:
$= 3 \cdot x \cdot \frac{4}{9}y^2$
Multiply the scalars ($3 \cdot \frac{4}{9} = \frac{12}{9}$), and reduce the fraction by dividing the numerator and denominator by their greatest common divisor, $3$:
$= \frac{4}{3}xy^2$ -
Fourth Term ($-b^3$):
$-(\frac{2}{3}y)^3$
Cube both the coefficient and the variable [$(\frac{2}{3})^3 = \frac{2^3}{3^3} = \frac{8}{27}$]:
$= -\frac{8}{27}y^3$
Step 3: Final Assembly of the Polynomial
Combine the simplified terms from Step 2 in descending order of the degree of $x$ to form the final expanded polynomial:
$x^3 - 2x^2y + \frac{4}{3}xy^2 - \frac{8}{27}y^3$
Final Solution: The expanded form of $(x - \frac{2}{3}y)^3$ is $x^3 - 2x^2y + \frac{4}{3}xy^2 - \frac{8}{27}y^3$.
More Questions from Class 9 Mathematics Polynomials EXERCISE 2.4
- Q1(i): Use suitable identities to find the following products: (i) $(x + 4) (x + 10)$
- Q1(ii): Use suitable identities to find the following products: (ii) $(x + 8) (x – 10)$
- Q1(iii): Use suitable identities to find the following products: (iii) $(3x + 4) (3x – 5)$
- Q1(iv): Use suitable identities to find the following products: (iv) $(y^2 + \frac{3}{2}) (y^2 – \frac{3}{2})$
- Q1(v): Use suitable identities to find the following products: (v) $(3 – 2x) (3 + 2x)$
- Q2(i): Evaluate the following products without multiplying directly: (i) $103 \times 107$
- Q2(ii): Evaluate the following products without multiplying directly: (ii) $95 \times 96$
- Q2(iii): Evaluate the following products without multiplying directly: (iii) $104 \times 96$
- Q3(i): Factorise the following using appropriate identities: (i) $9x^2 + 6xy + y^2$
- Q3(ii): Factorise the following using appropriate identities: (ii) $4y^2 – 4y + 1$
- Q3(iii): Factorise the following using appropriate identities: (iii) $x^2 – \frac{y^2}{100}$
- Q4(i): Expand each of the following, using suitable identities: (i) $(x + 2y + 4z)^2$
- Q4(ii): Expand each of the following, using suitable identities: (ii) $(2x – y + z)^2$
- Q4(iii): Expand each of the following, using suitable identities: (iii) $(–2x + 3y + 2z)^2$
- Q4(iv): Expand each of the following, using suitable identities: (iv) $(3a – 7b – c)^2$
- Q4(v): Expand each of the following, using suitable identities: (v) $(–2x + 5y – 3z)^2$
- Q4(vi): Expand each of the following, using suitable identities: (vi) $(\frac{1}{4}a - \frac{1}{2}b + 1)^2$
- Q5(i): Factorise: (i) $4x^2 + 9y^2 + 16z^2 + 12xy – 24yz – 16xz$
- Q5(ii): Factorise: (ii) $2x^2 + y^2 + 8z^2 – 2\sqrt{2}xy + 4\sqrt{2}yz – 8xz$
- Q6(i): Write the following cubes in expanded form: (i) $(2x + 1)^3$
- Q6(ii): Write the following cubes in expanded form: (ii) $(2a – 3b)^3$
- Q6(iii): Write the following cubes in expanded form: (iii) $(\frac{3}{2}x + 1)^3$
- Q7(i): Evaluate the following using suitable identities: (i) $(99)^3$
- Q7(ii): Evaluate the following using suitable identities: (ii) $(102)^3$
- Q7(iii): Evaluate the following using suitable identities: (iii) $(998)^3$
- Q8(i): Factorise each of the following: (i) $8a^3 + b^3 + 12a^2b + 6ab^2$
- Q8(ii): Factorise each of the following: (ii) $8a^3 – b^3 – 12a^2b + 6ab^2$
- Q8(iii): Factorise each of the following: (iii) $27 – 125a^3 – 135a + 225a^2$
- Q8(iv): Factorise each of the following: (iv) $64a^3 – 27b^3 – 144a^2b + 108ab^2$
- Q8(v): Factorise each of the following: (v) $27p^3 – \frac{1}{216} – \frac{9}{2}p^2 + \frac{1}{4}p$
- Q9(i): Verify : (i) $x^3 + y^3 = (x + y) (x^2 – xy + y^2)$
- Q9(ii): Verify : (ii) $x^3 – y^3 = (x – y) (x^2 + xy + y^2)$
- Q10(i): Factorise each of the following: (i) $27y^3 + 125z^3$ [Hint : See Question 9.]
- Q10(ii): Factorise each of the following: (ii) $64m^3 – 343n^3$ [Hint : See Question 9.]
- Q11: Factorise : $27x^3 + y^3 + z^3 – 9xyz$
- Q12: Verify that $x^3 + y^3 + z^3 – 3xyz = \frac{1}{2}(x + y + z)[(x – y)^2 + (y – z)^2 + (z – x)^2]$
- Q13: If $x + y + z = 0$, show that $x^3 + y^3 + z^3 = 3xyz$.
- Q14(i): Without actually calculating the cubes, find the value of each of the following: (i) $(–12)^3 + (7)^3 + (5)^3$
- Q14(ii): Without actually calculating the cubes, find the value of each of the following: (ii) $(28)^3 + (–15)^3 + (–13)^3$
- Q15(i): Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: (i) Area : $25a^2 – 35a + 12$
- Q15(ii): Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: (ii) Area : $35y^2 + 13y –12$
- Q16(i): What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (i) Volume : $3x^2 – 12x$
- Q16(ii): What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (ii) Volume : $12ky^2 + 8ky – 20k$
CBSE Solutions for Class 9 Mathematics Polynomials
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