Exploring Algebraic Identities Class 9 Extra Questions Maths Chapter 4 - #NCSOLVE 📚

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Class 9 Exploring Algebraic Identities Extra Questions

Extra Questions of Exploring Algebraic Identities

Class 9 Maths Chapter 4 Extra Questions

Exploring Algebraic Identities Class 9 Short Question Answer

Question 1.
Factorise x² – x – 12.
Solution:
We have,
x² – x – 12 = x² – 4x + 3x – 12
= x(x – 4) + 3(x – 4)
= (x – 4)(x + 3)
Thus, x² – x – 12 = (x – 4)(x + 3)

Question 2.
Factorise: 9a² – 9b² + 6a + 1
Solution:
9a² – 9b² + 6a + 1
= [9a² + 6a + 1] – 9b²
= [(3a)² + 2(3a)(1) + (1)²] – (3b)²
= (3a + 1)² – (3b)² = [(3a + 1) + 3b] [(3a + 1) – 3b]
[a² – b² = (a – b)(a + b)]
= (3a + 1 + 3b)(3a + 1 – 3b)

Question 3.
If x² – \(\frac{1}{x^2}\) = 18 then find the value of x – \(\frac{1}{x}\).
Solution:
(x – \(\frac{1}{x}\))² = \(\frac{1}{x^2}\) + – 2(x)(\(\frac{1}{x}\)) = x2 + \(\frac{1}{x^2}\) – 2
⇒ (x – \(\frac{1}{x}\))² = 18 – 2 = 16 = (4)²
⇒ (x – \(\frac{1}{x}\)) = ±4

Question 4.
Factorise : 14x⁶ – 45x³y³ – 14y⁶
Solution:
Let us put x³ = a and y³ = b so that
14x⁶ – 45x³y³ – 14y⁶ = 14a² – 45ab – 14b²
= 14a² – 49ab + 4ab — 14b²
= (2a – 7b) (7a + 2b)
⇒ 14x⁶ – 45x³y³ – 14y⁶ = (2x³ – 7y³) (7x³ + 2y³)

Question 5.
Find the product: (x – 3y) (x + 3y) (x² + 9y²)
Solution:
(x – 3y) (x + 3y) = (x² – 9y²)
and (x² – 9y) (x² + 9y)
= (x⁴ – 81y⁴)

Question 6.
If x + \(\frac{1}{2x}\) = 5, then find the value of x² + \(\frac{1}{4 x^2}\).
Solution:
We have x + \(\frac{1}{2x}\) = 5
[x + \(\frac{1}{2x}\)]² = 5²
⇒ x² + (\(\frac{1}{2x}\))² + 2 × x × \(\frac{1}{2x}\) = 25
⇒ x² + \(\frac{1}{4 x^2}\) + 1 = 25
⇒ x² + \(\frac{1}{4 x^2}\) = 25 – 1 = 24
Thus, the required value of x² + \(\frac{1}{4 x^2}\) is 24.

Question 7.
Factorise: x² + \(\frac{1}{x^3}\) – 2x – \(\frac{2}{x}\) + 2
Solution:
We have x² + \(\frac{1}{x^3}\) – 2x – \(\frac{2}{x}\) + 2

Question 8.
Show that x³ + y³ = (x + y )(x² – xy + y²).
Solution:
Since (x + y)³ = x³ + y³ + 3xy(x + y)
∴ x³ + y³ = (x + y)³ – 3xy(x + y)
= [(x + y)(x + y)] – 3xy(x + y)
= (x + y)[(x + y)² – 3xy]
= (x + y)[(x² + y² + 2xy) – 3xy]
= (x + y)[x² + y² – xy]
Thus, x³ + y³ = (x + y)(x² – xy + y²)

Question 9.
Show that x³ – y³ = (x – y)(x³ + xy + y³).
Solution:
Since (x – y)³ = x³ – y³ – 3xy(x – y)
∴ x³ – y³ = (x – y)³ + 3xy(x – y)
= [(x – y)(x – y)] + 3xy(x – y)
= (x – y)[(x – y)³ + 3xy]
= (x – y)[(x² + y² – 2xy) + 3xy]
= (x – y)[x² + y² – 2xy + 3xy]
= (x – y)(x² + y² + xy)
Thus, x³ – y³ = (x – y)(x² + xy + y²)

Question 10.
If x + y = 12 and xy = 27, find the value of x³ + y³.
Solution:
Since (x + y)³ = x³ + y³ + 3xy (x + y)
Substituting x + y = 12 and xy = 27, we have:
(12)³ = x³ + y³ + 3 (27) (12)
⇒ x³ + y³ = 123 – 81(12)
= [12³ – 9³] × 12
= [(12 + 9) (12 – 9)] × 12
= 21 × 3 × 12
= 756

Question 11.
Factorise: 8P³ + \(\frac{12}{5}\)P² + \(\frac{6}{25}\)P + \(\frac{1}{125}\)
Solution:
8P³ = (2P)³, and \(\frac{1}{125}=\left(\frac{1}{5}\right)^3\)
∴ 8P³ + \(\frac{12}{5}\)P² + \(\frac{6}{25}\)P + \(\frac{1}{125}\)
= (2P)³ + 3(2P)²(\(\frac{1}{5}\)) + 3(2P)(\(\frac{1}{5}\))² + (\(\frac{1}{2}\))³
= [2p + \(\frac{1}{5}\)]³
[a³ + 3a²b + 3ab² + b³ = (a + b)³]
= (2p + \(\frac{1}{5}\)) (2p + \(\frac{1}{5}\)) (2p + \(\frac{1}{5}\))

Question 12.
If p(x) = x² – 2√2 + 1, then find p(2√2).
Solution:
Since p(x) = x² – 2√2 + 1
Then p(2√2) = (2√2)² – (2√2)(2√2) + 1
= 4(2) – 4(2) + 1
= 8 – 8 + 1
= 1

Exploring Algebraic Identities Class 9 Long Question Answer

Question 1.
Factorise: x² – \(\frac{5}{12}\)x + \(\frac{1}{24}\)
Solution:

Thus, x² – \(\frac{5}{12}\)x + \(\frac{1}{24}\) = \(\frac{1}{24}\)(4x – 1)(6x – 1)

Question 2.
Factorise: (x⁶ – y⁶)
Solution:
x⁶ – y⁶ = (x³)² – (y³)²
= (x³ – y³)(x³ + y³) [v a² – b² = (a + b)(a – b)]
= [(x – y)(x² + xy + y²)] [(x + y)(x² – xy + y²)]
[∵ a³ + b³ = (a² + b² – ab)(a + b) and a³ – b³ = (a² + ab + b²)(a – b)]
= (x – y)(x + y)(x²+ xy + y²)(x² – xy + y²)
Thus, x⁶ – y⁶ = (x – y)(x + y)(x² + xy + y²)(x² – xy + y²)

Question 3.
Prove that (a + b + c)³ – a³ – b³ – c³ = 3(a + b) (b + c) (c + a)
Solution:
LHS= (a + b + c)³ – a³ – b³ – c³
= [(a + b + c)³ – a³] – [b³ + c³] …(1)
Now, (a + b + c)³ – a³
= (b + c) [3a² + b² + c² + 3ab + 2bc + 3ca] …(2)
[using x³ – y³ = (x – y) (x² + y² + xy)]
and b³ + c³ = (b + c) [b² + c² – bc] …(3)
[x³ + y³ = (x + y) (x² + y² -xy)]
From (1), (2) and (3), we get
LHS = (b + c) (3a³ + b² + c² + 3ab + 2bc + 3ca) – (b + c) (b² + c² – bc)
= (b + c) [3a² + b² + c² + 3ab + 2bc + 3ca – b² – c² + bc]
= (b + c) [3a² + (b² – b²) + (c² – c²) + 3ab + (2bc + bc) + 3ca]
= (b + c) [3a² + 0 + 0 + 3ab + 3bc + 3ca]
= (b + c) [3 (ab + ab + bc + ca)]
= 3(b + c) [(a + b) (c + a)]
= 3(a + b) (b + c) (c + a) = RHS

Question 4.
A group of (a + b) teachers, (a² + b²) girls and (a³ + b³) boys set out for an ‘Adult Education Mission’. If there are 10 teachers and 58 girls in the group, find the number of boys.
Solution:
∵ (a + b)² = a² + b² + 2ab
10² = 58 + 2ab
[∵ Number of teachers = (a + b) = 10
and Number of girls = (a² + b²) = 58]
⇒ 100 = 58 + 2ab
⇒ 2ab = 100 – 58 = 42
⇒ ab = \(\frac{42}{2}\) = 21
Now, (a + b)³ = a³ + b³ + 3 ab (a + b)
∴ (10)³ = a³ + b³ + 3 × 21 × 10
⇒ 1000 = a³ + b³ + 630
a³ + b³ = 1000 – 630 = 370
Number of boys = 370.

Exploring Algebraic Identities Class 9 Case Based Questions

Question 1.
A municipal corporation is redesigning a square-shaped city park that currently measures 100 m × 100 m. To improve functionality, they decide to expand the park by adding 5 m to both the length and the width, making it a larger square. Later, to build a footpath around it, they take 2 m off the length and 2 m off the width of this expanded park, making it (105 – 2) × (105 – 2).
Using this information, answer the following questions:
(i) Which algebraic identity is most useful to calculate the area of the initially expanded park (105 m × 105 m)?
(ii) What is the exact area of the expanded park (105 m × 105 m) using identities?
(iii) If the park dimensions become (105 – 2) × (105 – 2), find its area using a suitable identity.
Solution:
(i) The original park measures 100 m × 100 m.
After adding 5 m to each side, the new side length is:
105 = 100 + 5
So the area is (100 + 5)²
The most useful identity is (a + b)² = a² + 2ab + b²

(ii) Applying the identity:
(100 + 5)² = 100² + 2 × 100 × 5 + 5²
= 10000+ 1000 + 25
= 11025 m²
Thus, the area of the expanded park is 11,025 m².

(iii) Now, the side becomes 105 – 2 = 103
So, the area is: (103)² = (105 – 2)²
The suitable identity is (a – b)² = a² – 2ab + b²
Therefore, (105 – 2)² = 105² – 2 × 105 × 2 + 2²
= 11025 – 420 + 4
= 10609 m²
Thus, the area of the park is 10,609 m².

Exploring Algebraic Identities Class 9 Competency Based Questions

Question 1.
Ravi’s garden has dimensions (a + b) metres by (a – b) metres. If he reduces both the length and width by 1 metre, what will the new area be?
(a) (a² – b² – 2a – 2b + 1) m²
(b) (a² -b² – 2a – 2b – 1) m²
(c) (a² – b² – 2a – 2b) m²
(d) (a² – b² – 2a + 2b) m²
Solution:
(a) The new dimensions are (a + b – 1) and (a – b – 1). So, new area = (a + b – 1)(a – b – 1)
Expanding the product, we get
(a + b – 1)(a – b – 1) = a² – b² – 2a – 2b + 1
So, the new area will be (a² – b² – 2a – 2b + 1) m².

Question 2.
What is the value of (a + b)² – (a – b)²?
(a) 4ab
(b) 4a² + 4b²
(c) 2a² – 2b²
(d) 4a² – 4b²
Solution:
(a) (a + b)² – (a – b)² = [(a + b) – (a- b)][(a + b) + (a- b)]
On simplifying, we get
(2b)(2a) = 4 ab

Question 3.
Without actually calculating the cubes, find the value of each of the following:
(i) (- 12)³ + (7)³ + (5)³
(ii) (28)³ + (- 15)³ + (- 13)³
Solution:
(i) We have -12 + 7 + 5= 0
(- 12)³ + (7)³ + (5)³ = 3(— 12) (7)(5)
[Using the identity: If x + y + z = 0, then x³ + y³ + z³ = 3xyz]
= -36 × 35
=- 1260

(ii) We have 28 + (- 15) + (- 13) = 28 – 15 – 13 = 0
(28)³ + (- 15)³ + (- 13)³ = 3(28) (- 15) (- 13)
[Using the identity : If x + y + z = 0, then x³ + y³ + z³ = 3xyz] = 84 × 195 = 16380

Question 4.
If a, b, c are all non-zero and a + b + c = 0, prove that \(\frac{a^2}{b c}+\frac{b^2}{c a}+\frac{c^2}{a b}\).
Solution:
Since, a + b + c = 0,
a³ + b³ + c³ = 3abc
Now, in \(\frac{a^2}{b c}+\frac{b^2}{c a}+\frac{c^2}{a b}\) = 3, we have
LHS = \(\frac{a^2}{b c}+\frac{b^2}{c a}+\frac{c^2}{a b}\)
= \(\frac{a b c}{a b c}\left[\frac{a^2}{b c}+\frac{b^2}{c a}+\frac{c^2}{a b}\right]\) [Multiplying and dividing by ‘abc’]
= \(\frac{a^3+b^3+c^3}{a b c}\)

From (1) and (2), we have
LHS = \(\frac{3 a b c}{a b c}\) = 3 = RHS
∴ \(\frac{a^2}{b c}+\frac{b^2}{c a}+\frac{c^2}{a b}\) = 3

Question 5.
Ravita donated ₹ (x³ + \(\frac{1}{x^3}\)) to a blind school. Her friends wanted to know the amount donated by her. She did not disclose the amount but gave a hint that x + \(\frac{1}{x}\) = ₹ 7. Find the amount donated by Ravita to the blind school.
Solution:
x + \(\frac{1}{x}\) = 7 (Given)

Thus, the amount donated by Ravita is ₹ 322.

Exploring Algebraic Identities Extra Questions Class 9

Question 1.
Expand by using suitable identities.
(i) (2a + 3b)²
(ii) (4x²y + 6z)²
Solution:
(i) (2a + 3b)² = (2a)² + 2(2a)(3b) + (3b)² = 4a² + 12ab + 9b²
(ii) On comparing (4x²y + 6z)² with (a + b)², we get
a = 4x²y and b = 6z
By using the identity (a + b)² = a² + 2ab + b², we get
(4x²y + 6z)² = (4x²y)² + 2(4x²y) (6z) + (6z)² = 16x⁴y² + 48x²yz + 36z²

Question 2.
Expand the following by using suitable identities.
\(\left(2 x-\frac{1}{2 x}\right)^2\)
Solution:
On comparing \(\left(2 x-\frac{1}{2 x}\right)^2\) with (a – b)², we get
a = 2x and b = \(\frac {1}{2x}\)
By using identity (a – b)² = a² – 2ab + b², we get
\(\left(2 x-\frac{1}{2 x}\right)^2=(2 x)^2-2 \times 2 x \times \frac{1}{2 x}+\left(\frac{1}{2 x}\right)^2\) = \(4 x^2-2+\frac{1}{4 x^2}\)

Question 3.
Using a suitable identity, evaluate
(i) (1008)²
(ii) (9.98)²
Solution:
(i) Consider (1008)² = (1000 + 8)²
= (1000)² + (8)² + 2(1000)(8) [∵ (a + b)² = a² + b² + 2ab]
= 1000000 + 64 + 16000
= 1016064
(ii) Consider (9.98)² = (10 – 0.02)²
= (10)² + (0.02)² – 2(10)(0.02) [∵ (a – b)² = a² + b² – 2ab]
= 100 + 0.0004 – 0.4
= 99.6004

Question 4.
Evaluate 105 × 106 without multiplying directly.
Solution:
First, write 105 as 100 + 5 and 106 as 100 + 6.
Then, 105 × 106 = (100 + 5)(100 + 6)
= (100)² + (5 + 6) (100) + (5 × 6) [∵ (x + a)(x + b) = x² + (a + b)x + ab]
= 10000 + 11 × 100 + 30
= 10000 + 1100 + 30
= 11130

Question 5.
If x ≠ 0, x + \(\frac {1}{2x}\) = p and x – \(\frac {1}{2x}\) = q then find a relation between p and q.
Solution:
We have, x + \(\frac {1}{2x}\) = p …..(i)
and x – \(\frac {1}{2x}\) = q …..(ii)
On squaring both sides of Eq. (i), we get
\(\left(x+\frac{1}{2 x}\right)^2=p^2\)
⇒ \(x^2+\frac{1}{4 x^2}+2 \times x \times \frac{1}{2 x}=p^2\) [∵ (a + b)² = a² + b² + 2ab]
⇒ \(x^2+\frac{1}{4 x^2}+1=p^2\)
⇒ \(x^2+\frac{1}{4 x^2}=p^2-1\) ……(iii)
Now, squaring both sides of Eq. (ii), we get
\(\left(x-\frac{1}{2 x}\right)^2=q^2\)
⇒ \(x^2+\frac{1}{4 x^2}-2 \times x \times \frac{1}{2 x}=q^2\) [∵ (a – b)² = a² + b² – 2ab]
⇒ \(x^2+\frac{1}{4 x^2}-1=q^2\)
⇒ \(x^2+\frac{1}{4 x^2}=q^2+1\) …….(iv)
From Eqs. (iii) and (iv), we get
p² – 1 = q² + 1
⇒ p² – q² = 2

Question 6.
If a + b = 6 and ab = 8, then find the value of (a – b).
Solution:
Consider (a – b)² = (a + b)² – 4ab
⇒ (a – b)² = 6² – 4 × 8 [∵ a + b = 6 and ab = 8]
⇒ (a – b)² = 36 – 32 = 4
⇒ (a – b)² = 4
⇒ a – b = ±2 [taking square root on both sides]

Question 7.
If a² – 4a – 1 = 0 and a ≠ 0, then find the values of
(i) a – \(\frac {1}{a}\)
(ii) a + \(\frac {1}{a}\)
(iii) \(a^2-\frac{1}{a^2}\)
Solution:

Question 8.
If \(\frac{a}{b}=\frac{b}{c}\) then prove that (a + b + c)(a – b + c) = a² + b² + c².
Solution:
We have, \(\frac{a}{b}=\frac{b}{c}\) = k (say)
⇒ a = bk and b = ck
⇒ a = ck²
Consider LHS = (a + b + c)(a – b + c)
= (ck² + ck + c)(ck² – ck + c) [∵ a = ck² and b = ck]
= c² (k² + k + 1) (k² – k + 1)
= c² [{(k² + 1) + k}{(k² + 1) – k}]
= c² {(k² + 1)² – k²} [∵ (a + b)(a – b) = a² – b²]
= c² [k⁴ + 1 + 2k² – k²] [∵ (a + b)² = a² + b² + 2ab]
= c² {k⁴ + k² + 1} …..(i)
Now, consider RHS = a² + b² + c²
= (ck²)² + (ck)² + c² [∵ a = ck² and b = ck]
= c²k⁴ + c²k² + c²
= c² (k⁴ + k² + 1) ……(ii)
From Eqs. (i) and (ii), we get
LHS = RHS
Hence proved.

Question 9.
Write the following in the expanded form.
(i) (3x + y + 2)²
(ii) (2x – y + 4z)²
(iii) (4x – 2y – 3z)²
(iv) \(\left(\frac{2}{3} x-\frac{3}{4} y+\frac{4}{5} z\right)^2\)
Solution:
(i) We have, (3x + y + 2)²
On comparing with (a + b + c)², we get
a = 3x, b = y, and c = 2
We know that (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
On putting the values of a, b, and c in the above identity, we get
(3x + y + 2)² = (3x)² + (y)² + (2)² + 2 × (3x) × (y) + 2 × y × 2 + 2 × 2 × 3x = 9x² + y² + 4 + 6xy + 4y + 12x
(ii) Do the same as part (i).
(iii) Do the same as Part (i).
(iv) Do the same as Part (i).

Question 10.
Simplify (x + y + z)² + (x – y – z)².
Solution:
Consider (x + y + z)² + (x – y – z)²
= x² + y² + z² + 2xy + 2yz + 2zx + x² + y² + z² – 2xy + 2yz – 2zx
= 2x² + 2y² + 2z² + 4yz
= 2(x² + y² + z² + 2yz)

Question 11.
Find the cube of x – \(\frac {1}{x}\) + y, where x ≠ 0.
Solution:

Question 12.
Find the product of (x + y + 3z) (x² + y² + 9z² – xy – 3yz – 3xz).
Solution:
We have, (x + y + 3z) (x² + y² + 9z² – xy – 3yz – 3xz) = (x + y + 3z) [x² + y² + (3z)² – x × y – y × 3z – 3z × x]
On comparing it with (a + b + c)(a² + b² + c² – ab – bc – ca), we get
a = x, b = y, and c = 3z
Now, as (a + b + c)(a² + b² + c² – ab – bc – ca) = a³ + b² + c³ – 3abc
[∵ (x + y + 3z) (x² + y² + 9z² – xy – 3yz – 3xz)]
= [x³ + y³ + (3z)³ – 3 × x × y × 3z]
= x³ + y³ + 27z³ – 9xyz

Question 13.
If x = y + 5, then find the value of x³ – y³ – 15xy – 125.
Solution:
Given, x = y + 5
⇒ x – y – 5 = 0
⇒ x + (-y) + (-5) = 0
⇒ (x)³ + (-y)³ + (-5)³ = 3(x)(-y)(-5) [∵ If (a + b + c) = 0, then a³ + b³ + c³ = 3abc]
⇒ x³ – y³ – 125 = 15xy
⇒ x³ – y³ – 15xy – 125 = 0

Question 14.
If x + y + z = 0, then find the value of \(\frac{x^2}{y z}+\frac{y^2}{x z}+\frac{z^2}{x y}\).
Solution:
x + y + z = 0
∴ \(\frac{x^2}{y z}+\frac{y^2}{x z}+\frac{z^2}{x y}=\frac{x^3+y^3+z^3}{x y z}\) = \(\frac{3 x y z}{x y z}\) = 3
[∵ If x + y + z = 0, then x³ + y³ + z³ = 3xyz]

Question 15.
Factorise the following.
(i) 12x – 60
(ii) -18yz + 27y²z²
(iii) 6a²b + 9ab² + 3abc
Solution:
(i) In the given expression, we see that each term can be divided by 12.
So, 12 is the common term.
12x – 60 = 12\(\left(\frac{12 x}{12}-\frac{60}{12}\right)\) = 12(x – 5)
which are the required factors of the given expression.
(ii) Do the same as Part (i).
(iii) Do the same as Part (i)

Question 16.
Factorise the following.
(i) 12x(x + 3y)³ – 18y(x + 3y)² + 24(x + 3y)
(ii) 15a(a – 2b)³ – 10b(a – 2b)² + 20(a – 2b)
Solution:
(i) In the given expression, we see that each term can be divided by 6(x + 3y).
So, 6(x + 3y) is the common term.
= 6(x + 3y)\(\left[\begin{array}{r}
\frac{12 x(x+3 y)^3}{6(x+3 y)} \\
-\frac{18 y(x+3 y)^2}{6(x+3 y)} \\
+\frac{24(x+3 y)}{6(x+3 y)}
\end{array}\right]\)
= 6(x + 3y) [2x(x + 3y)2 – 3y(x + 3y) + 4),
which are the required factors of the given expression.
(ii) Do the same as Part (i).

Question 17.
Factorise each of the following expressions.
(i) 3x² – 27
(ii) 20x – 5x³
Solution:
(i) 3x² – 27 = 3(x² – 9)
= 3(x² – 3²)
= 3(x + 3)(x – 3)
(ii) 20x – 5x³ = 5x(4 – x²)
= 5x(2² – x²)
= 5x(2 + x)(2 – x)

Question 18.
Factorise each of the following expressions.
(i) \(x^2+\frac{1}{x^2}-5\)
(ii) \(x^2+\frac{1}{x^2}-6-y^2-\frac{1}{y^2}\)
Solution:

Question 19.
Find possible expressions for the length, breadth and height of each cuboid, whose volumes are given below (in cubic units).
(i) 10m² – 40n²
(ii) 2r²s + 6rs²
Solution:
(i) We have, 10m² – 40n² = 10(m² – 4n²) = 10[(m)² – (2n)²]
Using the identity a² – b² = (a – b)(a + b), we get
10m² – 40n² = 10(m – 2n)(m + 2n)
Hence, possible length = 10, breadth = m – 2n, and height = m + 2n
(ii) We have, 2r²s + 6rs² = 2rs(r + 3s) = (2r)(s)(r + 3s)
Hence, possible length = 2r, breadth = s, and height = r + 3s.

Question 20.
Factorise 36x² + 25y² + 49z² + 60xy – 70yz – 84xz.
Solution:
We have, 36x² + 25y² + 49z² + 60xy – 70yz – 84xz
= (6x)² + (5y)² + (-7z)² + 2(6x)(5y) + 2(5y)(-7z) + 2(-7z)(6x)
= [(6x) + (5y) + (-7z)]²
= (6x + 5y – 7z)²
= (6x + 5y – 7z)(6x + 5y – 7z)
These are the required factors of the given expression.

Question 21.
Factorise 27 – 125a³ – 135a + 225a².
Solution:
We have, 27 – 125a³ – 135a + 225a² = {(3)³ – (5a)³} – 45a(3 – 5a)
= (3 – 5a){(3)² + 3 × 5a + (5a)²} – 45a(3 – 5a)
=(3 – 5a)(9 + 15a + 25a²) – 45a(3 – 5a)
= (3 – 5a)(9 + 15a + 25a² – 45a)
= (3 – 5a){9 – 30a + 25a²}
= (3 – 5a){(3)² – 2 × 3 × 5a + (5a)²}
= (3 – 5a)(3 – 5a)²
= (3 – 5a)(3 – 5a) (3 – 5a)

Question 22.
Prove that a³ + b³ + c³ – 3abc = \(\frac {1}{2}\)(a + b + c)[(a – b)² + (b – c)² + (c – a)²].
Solution:
a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)
= \(\frac {1}{2}\)(a + b + c)(2a² + 2b² + 2c² – 2ab – 2bc – 2ca)
Now, rearranging the terms, we get
= \(\frac {1}{2}\)(a + b + c)[(a² – 2ab + b²) + (b² – 2ab + c²) + (c² – 2ca + a²)]
= \(\frac {1}{2}\)(a + b + c)[(a – b)² + (b – c)² + (c – a)²]

Question 23.
Factorise (x – 3y)³ + (3y – 4z)³ + (4z – x)³.
Solution:
Put x – 3y = a, 3y – 4z = b, 4z – x = c
∴ a + b + c = x – 3y + 3y – 4z + 4z – x = 0
⇒ a³ + b³ + c³ = 3abc
⇒ (x – 3y)³ + (3y – 4z)³ + (4z – x)³ = 3(x – 3y)(3y – 4z)(4z – x),
which are the required factorisations.

Question 24.
Simplify the following rational expressions.
(i) \(\frac{3 x^2-27}{x^2-9}\)
(ii) \(\frac{x^2-7 x+10}{x-5}\)
(iii) \(\frac{x^2-10 x+25}{x^2-25}\)
Solution:
(i) Factorise numerator and denominator:
3x² – 27 = 3(x² – 9) = 3(x – 3)(x + 3)
x² – 9 = (x – 3)(x + 3)
(ii) Factorise numerator
x² – 7x + 10 = (x – 5)(x – 2)
(iii) Factorise numerator and denominator
x² – 10x + 25 = (x – 5)²
x² – 25 = (x – 5)(x + 5)

Exploring Algebraic Identities Class 9 Extra Questions for Practice

Multiple Choice Questions

Question 1.
If 49x² – y = (7x + \(\frac{1}{2}\))(7x – \(\frac{1}{2}\)), then what is the value of y?
(a) 0
(b)
(c) 1
(d) 1

Question 2.
The value of 249² – 248² is
(a) 12
(b) 477
(c) 487
(d) 497

Question 3.
The factorisation of 4x² + 8x + 3 is
(a) (x + 1 )(x + 3)
(b) (2x + 1)(2x + 3)
(c) (x + 2)(2x + 5)
(d) (2x – 1) (2x – 3)

Assertion-Reason Questions

Two statements are given, one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer from the options (a), (b), (c), and (d) given below.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false, but Reason (R) is true.

Question 1.
Assertion (A): The expansion of (a + b)³ is a³ + b³ + 3ab{a + b).
Reason (R): This is the correct factorisation of the sum of cubes.

Question 2.
Assertion (A): The identity (a + b)² – (a – b)² = 4ab holds true for all values of a and b.
Reason (R): This is an application of the difference of squares identity.

Short Answer Type Questions

Question 1.
Without actually calculating the cubes, find the value of 48³ – 30³ – 18³.

Question 2.
Factorise: 4x² + 20x + 25.

Question 3.
Use suitable identity to find the product of (y² – \(\frac{3}{2}\))(y² – \(\frac{3}{2}\))

Question 4.
Evaluate the following using suitable identities: \(\frac{1}{2}\)

Question 5.
Using the identity for the square of a binomial, simplify the expression (x + 3)² – (x – 3)².

Question 6.
If the sum of a number and its reciprocal equals , find the number.

Long Answer Type Questions

Question 1.
Prove the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca using algebraic expansion.

Question 2.
A rectangular pool has area (2x² + 7x + 3) square ft. If its width is (2x + 1) ft, find its length.

The post Exploring Algebraic Identities Class 9 Extra Questions Maths Chapter 4 appeared first on Learn CBSE.

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