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Q11:
Factorise : $27x^3 + y^3 + z^3 – 9xyz$
Solution :
Step 1: Algebraic Setup and Identity Identification
We are tasked with factorising the following multivariable polynomial:
$27x^3 + y^3 + z^3 - 9xyz$
By observing the structure of the expression, we can see it consists of a sum of three perfect cubes followed by a product term. This specific structure corresponds directly to the standard algebraic identity for the sum of three cubes:
[Standard Algebraic Identity]
$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
Step 2: Term-by-Term Structural Mapping
To apply the identity, we must rewrite the given polynomial so that each term perfectly matches the $a^3$, $b^3$, $c^3$, and $-3abc$ components of the formula. We extract the cube roots of the first three terms:
- For the first term: $27x^3 = (3^3)(x^3) = (3x)^3$ [By the power of a product property: $x^n y^n = (xy)^n$]
- For the second term: $y^3 = (y)^3$
- For the third term: $z^3 = (z)^3$
Now, we verify if the fourth term, $-9xyz$, matches the $-3abc$ structure using our derived values for $a$, $b$, and $c$:
$-3(3x)(y)(z) = -9xyz$
Since the product matches the original expression exactly, we can definitively establish our variable mappings:
Step 3: Substitution into the Algebraic Identity
We now substitute $a = 3x$, $b = y$, and $c = z$ into the right-hand side of our identity:
$(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
Substituting the mapped variables yields:
$= (3x + y + z) \left[ (3x)^2 + (y)^2 + (z)^2 - (3x)(y) - (y)(z) - (z)(3x) \right]$
Step 4: Simplification of the Polynomial Factors
To finalize the factorisation, we must expand and simplify the terms within the second bracket:
- Square the first term: $(3x)^2 = 9x^2$
- Square the second term: $(y)^2 = y^2$
- Square the third term: $(z)^2 = z^2$
- Compute the first cross-product: $-(3x)(y) = -3xy$
- Compute the second cross-product: $-(y)(z) = -yz$
- Compute the third cross-product: $-(z)(3x) = -3zx$
Reassembling the simplified terms into the second bracket, we get:
$(9x^2 + y^2 + z^2 - 3xy - yz - 3zx)$
Combining this with the first bracket gives us the complete, fully factorised expression.
Final Solution: The factorised form of the polynomial $27x^3 + y^3 + z^3 - 9xyz$ is $(3x + y + z)(9x^2 + y^2 + z^2 - 3xy - yz - 3zx)$.
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) $(3 – 2x) (3 + 2x)$
- 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.]
- 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 : $12ky^2 + 8ky – 20k$
- 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) $(\frac{3}{2}x + 1)^3$
- Q6(iv): Write the following cubes in expanded form: (iv) $(x - \frac{2}{3}y)^3$
- 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) $x^3 – y^3 = (x – y) (x^2 + xy + y^2)$
CBSE Solutions for Class 9 Mathematics Polynomials
Chapters in CBSE - Class 9 Mathematics
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