NCERT solution class 8 chapter 9 Algebraic Expression and Identities exercise 9.3

Exercise 9.3

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Q 1 - Ex 9.3 - Algebraic Expressions and Identities - NCERT Maths Class 8th - Chapter 9

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Question 1
Carry out the multiplication of the expressions in each of the following pairs.
(i) 4p, q + r (ii) ab, a − b (iii) a + b, 7a²b²
(iv) a² − 9, 4a (v) pq + qr + rp, 0

Sol :
(i) (4p) × (q + r) = (4p × q) + (4p × r) = 4pq + 4pr
(ii) (ab) × (a − b) = (ab × a) + [ab × (− b)] = a²b − ab²
(iii) (a + b) × (7a² b²) = (a × 7a²b²) + (b × 7a²b²) = 7a³b² + 7a²b³
(iv) (a² − 9) × (4a) = (a² × 4a) + (− 9) × (4a) = 4a³ − 36a
(v) (pq + qr + rp) × 0 = (pq × 0) + (qr × 0) + (rp × 0) = 0

Q 2 - Ex 9.3 - Algebraic Expressions and Identities - NCERT Maths Class 8th - Chapter 9

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Question 2
Complete the table

---	First expression	Second Expression	Product
(i)	a	b + c + d	-
(ii)	x + y − 5	5 xy	-
(iii)	p	6p2 − 7p + 5	-
(iv)	4p2q2	p2 − q2	-
(v)	a + b + c	abc	-

Sol :

The table can be completed as follows.

-	First expression	Second Expression	Product
(i)	a	b + c + d	ab + ac + ad
(ii)	x + y − 5	5 xy	5x2y + 5xy2 − 25xy
(iii)	p	6p2 − 7p + 5	6p3 − 7p2 + 5p
(iv)	4p2q2	p2 − q2	4p4q2 − 4p2q4
(v)	a + b + c	abc	a2bc + ab2c + abc2

Q 3 - Ex 9.3 - Algebraic Expressions and Identities - NCERT Maths Class 8th - Chapter 9

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Question 3
Find the product.
(i) (a²) × (2a²²) × (4a²⁶)
(ii)
(iii)
(iv) x × x² × x³ × x⁴

Sol :
(i) (a²) × (2a²²) × (4a²⁶) = 2 × 4 ×a² × a²² × a²⁶ = 8a⁵⁰
(ii)
(iii)
(iv) x × x² × x³ × x⁴ = x¹⁰

Q 4 - Ex 9.3 - Algebraic Expressions and Identities - NCERT Maths Class 8th - Chapter 9

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Question 4
(a) Simplify 3x (4x −5) + 3 and find its values for (i) x = 3, (ii) .
(b) a (a² + a + 1) + 5 and find its values for (i) a = 0, (ii) a = 1, (iii) a = − 1.

Sol :
(a) 3x (4x − 5) + 3 = 12x² − 15x + 3
(i) For x = 3, 12x² − 15x + 3 = 12 (3)² − 15(3) + 3
= 108 − 45 + 3
= 66
(ii) For

(b)a (a² + a + 1) + 5 = a³ + a² + a + 5
(i) For a = 0, a³ + a² + a + 5 = 0 + 0 + 0 + 5 = 5
(ii) For a = 1, a³ + a² + a + 5 = (1)³ + (1)² + 1 + 5
= 1 + 1 + 1 + 5 = 8
(iii) For a = −1, a³ + a² + a + 5 = (−1)³ + (−1)² + (−1) + 5
= − 1 + 1 − 1 + 5 = 4



Question 5
(a) Add: p (p − q), q (q _­­­− r) and r (r ­− p)
(b) Add: 2x (z − x − y) and 2y (z − y − x)
(c) Subtract: 3l (l − 4m + 5n) from 4l (10n − 3m + 2l)
(d) Subtract: 3a (a + b + c) − 2b (a − b + c) from 4c (− a + b + c)

Sol :
(a) First expression = p (p − q) = p² − pq
Second expression = q (q _­­­− r) = q² − qr
Third expression = r (r ­− p) = r² − pr
Adding the three expressions, we obtain

Therefore, the sum of the given expressions is p² + q² + r² − pq − qr − rp.
(b) First expression = 2x (z − x − y) = 2xz − 2x² − 2xy
Second expression = 2y (z − y − x) = 2yz − 2y² − 2yx
Adding the two expressions, we obtain

Therefore, the sum of the given expressions is − 2x² − 2y² − 4xy + 2yz + 2zx.
(c) 3l (l − 4m + 5n) = 3l² − 12lm + 15ln
4l (10n − 3m + 2l) = 40ln − 12lm + 8l²
Subtracting these expressions, we obtain

Therefore, the result is 5l² + 25ln.
(d) 3a (a + b + c) − 2b (a − b + c) = 3a² +3ab + 3ac − 2ba + 2b² − 2bc
= 3a² + 2b² + ab + 3ac − 2bc
4c (− a + b + c) = − 4ac + 4bc + 4c²
Subtracting these expressions, we obtain

Therefore, the result is −3a² −2b² + 4c² − ab + 6bc − 7ac.