The Class 8 is an important year in a student’s life and Maharashtra State Board Maths is one of the subjects that require dedication, hard work, and practice. It’s a subject where you can score well if you are well-versed with the concepts, remember the important formulas and solving methods, and have done an ample amount of practice. Worry not! Home Revise is here to make your Class 8 journey even easier. It’s essential for students to have the right study material and notes to prepare for their board examinations, and through Home Revise, you can cover all the fundamental topics in the subject and the complete Maharashtra State Board Class 8 Maths Book syllabus.
Practice set 5.1 PAGE NO: 24
1. Expand:
Solution:
(1) (a + 2) (a – 1)
Let us simplify the expression, we get
(a + 2) (a – 1) = a2 + [(2) + (-1)] a + [(2) × (-1)]
By using the logic,
(x + p) (x + q) = x2 + (p + q)x +(p × q)
Here, x = a, p = 2, q = -1
Now, substitute the value we get
= a2 + (2 – 1)a + (-2)
= a2 + 2a – a – 2
= a2 + a – 2
∴ (a + 2) (a – 1) = a2 + a – 2
(2) (m – 4) (m + 6)
Let us simplify the expression, we get
(m – 4) (m + 6) = m2 + [(- 4) + (6)] m + [(- 4) × (6)]
By using the logic,
(x + p) (x + q) = x2 + (p + q)x + (p × q)
Here, x = m, p = -4, q = 6
Now, substitute the value we get
= m2 + (6 – 4)m + (- 24)
= m2 + 6m – 4m – 24
= m2 + 2m – 24
∴ (m – 4) (m + 6) = m2 + 2m – 24
(3) (p + 8) (p – 3)
Let us simplify the expression, we get
(p + 8) (p – 3) = p2 + [(8) + (- 3)] p + [(8) × (- 3)]
By using the logic,
(x + a) (x + b) = x2 + (a + b)x +(a × b)
Here, x = p, a = 8, b = -3
Now, substitute the value we get
= p2 + (8 – 3)p + (- 24)
= p2 + 8p – 3p – 24
= p2 + 5p – 24
∴ (p + 8) (p – 3) = p2 + 5p – 24
(4) (13 + x) (13 – x)
Let us simplify the expression, we get
(13 + x) (13 – x) = (13)2 – (x)2
{We know that (a + b) (a – b) = (a)2 – (b)2 }
= 169 + 0(13) – x2
= 169 – x2
∴ (13 + x) (13 – x) = 169 – x2
(5) (3x + 4y) (3x + 5y)
Let us simplify the expression, we get
(3x + 4y) (3x + 5y) = (3x)2 + [(4y) + (5y)] 3x + [(4y) × (5y)]
By using the logic,
(x + a) (x + b) = x2 + (a + b)x + (a × b)
Here, x = 3x, a = 4y, b = 5y
Now, substitute the value we get
= 9x2 + [(9y) × (3x)] + 20y2
= 9x2 + 27xy + 20y2
∴ (3x + 4y) (3x + 5y) = 9x2 + 27xy + 20y2
(6) (9x – 5t) (9x + 3t)
Let us simplify the expression, we get
(9x – 5t) (9x + 3t) = (9x)2 + [(- 5t) + (3t)] 9x + [(- 5t) × (3t)]
By using the logic,
(x + a) (x + b) = x2 + (a + b)x + (a × b)
Here, x = 9x, a = -5t, b = 3t
Now, substitute the value we get
= 81x2 + [(- 2t) × (9x)] + (- 15t2 )
= 81x2 – 18xt – 15t2
∴ (9x – 5t) (9x + 3t) = 81x2 – 18xt – 15t2
Practice set 5.2 PAGE NO: 25
1. Expand:
Solution:
(1) (k + 4)3
Let us simplify the expression, we get
(k + 4)3 = (k)3 + [ 3 ×(k)2 × (4)] + [ 3 × (k) × (4)2 ] + (4)3
By using the formula,
(a + b)3 = a3 + 3a2 b + 3ab2 + b3
Here a = k, b = 4
Now, substitute the value we get
= k3 + (3 × 4)k2 + (3 × 16)k + 64
= k3 + 12k2 + 48k + 64
∴ (k + 4)3 = k3 + 12k2 + 48k + 64
(2) (7x + 8y)3
Let us simplify the expression, we get
(7x + 8y)3 = (7x)3 + [ 3 × (7x)2 × (8y)] + [ 3 × (7x) × (8y)2 ] + (8y)3
By using the formula,
(a + b)3 = a3 + 3a2 b + 3ab2 + b3
Here a = 7x, b = 8y
Now, substitute the value we get
= 343x3 + (3 × 49 × 8)x2 y + (3 × 7 × 64)xy2 + 512y3
= 343x3 + 1176x2 y + 1344xy2 + 512y3
∴ (7x + 8y)3 = 343x3 + 1176x2 y + 1344xy2 + 512y3
(3) (7 + m)3
Let us simplify the expression, we get
(7 + m)3 = (7)3 + [ 3 × (7)2 × (m)] + [ 3 × (7) × (m)2 ] + (m)3
By using the formula,
(a + b)3 = a3 + 3a2 b + 3ab2 + b3
Here a = 7, b = m
Now, substitute the value we get
= 343 + (3 × 49)m + (3 × 7)m2 + m3
= 343 + 147m + 21m2 + m3
∴ (7 + m)3 = 343 + 147m + 21m2 + m3
(4) (52)3
Let us simplify the expression, we get
(52)3 = (50 + 2)3
(50 + 2)3 = (50)3 + [3 × (50)2 × (2)] + [ 3 × (50) × (2)2 ] + (2)3
By using the formula,
(a + b)3 = a3 + 3a2 b + 3ab2 + b3
Here a = 50, b = 2
Now, substitute the value we get
= 125000 + (3 × 2500 × 2) + (3 × 50 × 4) + 8
= 125000 + 15000 + 600 + 8
= 140608
∴ (52)3 = (50 + 2)3 = 140608
(5) (101)3
Let us simplify the expression, we get
(101)3 = (100 + 1)3
(100 + 1)3 = (100)3 + [ 3 ×(100)2 ×(1)] + [ 3 ×(100)×(1)2 ] +(1)3
By using the formula,
(a + b)3 = a3 + 3a2 b + 3ab2 + b3
Here a = 100, b = 1
Now, substitute the value we get
= 1000000 + (3 × 10000 × 1) + (3 × 100 × 1) + 1
= 1000000 + 30000 + 300 + 1
= 1030301
∴ (101)3 = (100 + 1)3 = 1030301
Practice set 5.3 PAGE NO: 27
1. Expand:
Solution:
(1) (2m – 5)3
Let us simplify the expression, we get
(2m – 5)3 = (2m)3 – [ 3 × (2m)2 × 5 ] + [ 3 × (2m) × (5)2 ] – (5)3
By using the formula,
(a – b)3 = a3 – 3a2 b + 3ab2 – b3
Here, a = 2m, b = -5
Now, substitute the value we get
= 8m3 – [3 × 4m2 × 5] + [ 3 × 2m × 25] – 125
= 8m3 – 60m2 + 150m – 125
∴ (2m – 5)3 = 8m3 – 60m2 + 150m – 125
(2) (4 – p)3
Let us simplify the expression, we get
(4 – p)3 = (4)3 – [ 3 × (4)2 × p ] + [ 3 × (4) × (p)2 ] – (p)3
By using the formula,
(a – b)3 = a3 – 3a2 b + 3ab2 – b3
Here, a = 4, b = -p
Now, substitute the value we get
= 64 – [3 × 6 × p ] + [ 3 × 4 × p2 ] – p3
= 64 – 48p + 12p2 – p3
∴ (4 – p)3 = 64 – 48p + 12p2 – p3
(3) (7x – 9y)3
Let us simplify the expression, we get
(7x – 9y)3 = (7x)3 – [ 3 × (7x)2 × 9y ] + [3 × (7x) × (9y)2 ] – (9y)3
By using the formula,
(a – b)3 = a3 – 3a2 b + 3ab2 – b3
Here, a = 7x, b = -9y
Now, substitute the value we get
= 343x3 – [3 × 49x2 × 9y] + [3 × 7x × 81y2 ] – 729y3
= 343x3 – 1323x2 y + 1701xy2 – 729y3
∴ (7x – 9y)3 = 343x3 – 1323x2 y + 1701xy2 – 729y3
(4) (58)3
Let us simplify the expression, we get
(58)3 = (60 – 2)3
(60 – 2)3 = (60)3 – [3 × (60)2 × 2] + [3 × (60) × (2)2 ] – (2)3
By using the formula,
(a – b)3 = a3 – 3a2 b + 3ab2 – b3
Here, a = 60, b = -2
Now, substitute the value we get
= 216000 – [3 × 3600 × 2] + [3 × 60 × 4] – 8
= 216000 – 21600 + 720 – 8
= 195112
∴ (58)3 = (60 – 2)3 = 195112
(5) (198)3
Let us simplify the expression, we get
(198)3 = (200 – 2)3
(200 – 2)3 = (200)3 – [3 × (200)2 × 2] + [3 × (200) × (2)2 ] – (2)3
By using the formula,
(a – b)3 = a3 – 3a2 b + 3ab2 – b3
Here, a = 200, b = -2
Now, substitute the value we get
= 8000000 – 240000 + 2400 – 8
= 7762392
∴ (198)3 = (200 – 2)3 = 7762392
2. Simplify:
(1) (2a + b)3 – (2a – b)3
(2) (3r – 2k)3 + (3r + 2k)3
(3) (4a – 3)3 – (4a + 3)3
(4) (5x – 7y)3 + (5x + 7y)3
Solution:
(1) (2a + b)3 – (2a – b)3
Let us expand the given expression:
(2a + b)3 –(2a – b)3 = [(2a)3 +{3 ×(2a)2 × b } + {3 ×(2a)×(b)2 } +(b)3 ] –[(2a)3 -{3 × (2a)2 × b } +{3 ×(2a)×(b)2 } -(b)3 ]
By using the formula,
(a + b)3 = a3 + 3a2 b + 3ab2 + b3 and (a – b)3 = a3 – 3a2 b + 3ab2 – b3
= [8a3 + {3 × 4a2 × b} + {3 × 2a ×b} + b3 ] – [8a3 – {3 × 4a2 × b} + {3 × 2a × b2 } – b3 ]
= [8a3 + 12a2 b + 6ab2 + b3 ] – [8a3 – 12a2 b + 6ab2 – b3 ]
= 8a3 + 12a2 b + 6ab2 + b3 – 8a3 + 12a2 b – 6ab2 + b3
= 24a2 b + 2b3
∴ (2a + b)3 – (2a – b)3 = 24a2 b + 2b3
(2) (3r – 2k)3 + (3r + 2k)3
Let us expand the given expression:
(3r – 2k)3 + (3r + 2k)3 = [(3r)3 -{3 ×(3r)2 ×(2k)} + {3 ×(3r)×(2k)2 } -(2k)3 ] + [(3r)3 +{3 × (3r)2 ×(2k)} + {3 ×(3r)×(2k)2 } +(2k)3 ]
By using the formula,
(a + b)3 = a3 + 3a2 b + 3ab2 + b3 and (a – b)3 = a3 – 3a2 b + 3ab2 – b3
= [27r3 – {3 × 9r2 × 2k} + {3 × 3r × 4k2 } – 8k3 ] + [27r3 + {3 × 9r2 × 2k} + {3 × 3r ×(4k2 )} + 8k3 ]
= [27r3 – 54r2 k + 36rk2 – 8k3 ] + [27r3 + 54r2 k + 36rk2 + 8k3 ]
= 27r3 –54r2 k + 36rk2 – 8k3 + 27r3 + 54r2 k + 36rk2 + 8k3
= 54r3 + 72rk2
∴ (3r – 2k)3 + (3r + 2k)3 = 54r3 + 72rk2
(3) (4a – 3)3 – (4a + 3)3
Let us expand the given expression:
(4a – 3)3 – (4a + 3)3 = [(4a)3 –{3 ×(4a)2 × 3 } + {3 ×(4a)×(3)2 } –(3)3 ] –[(4a)3 +{3 × (4a)2 × 3} + {3 ×(4a)×(3)2 } + (3)3 ]
By using the formula,
(a + b)3 = a3 + 3a2 b + 3ab2 + b3 and (a – b)3 = a3 – 3a2 b + 3ab2 – b3
= [64a3 –{3 × 16a2 × 3} + {3 × 4a × 9} –27] – [64a3 + {3 × 16a2 × 3} + {3 × 4a × 9} + 27]
= [64a3 –144a2 + 108a –27] – [64a3 + 144a2 + 108a + 27]
= 64a3 –144a2 + 108a – 27 – 64a3 –144a2 –108a –27
= –288a2 – 54
∴ (4a – 3)3 – (4a + 3)3 = -288a2 – 54
(4) (5x – 7y)3 + (5x + 7y)3
Let us expand the given expression:
(5x – 7y)3 + (5x + 7y)3 = [(5x)3 -{3 ×(5x)2 × (7y)} +{3 ×(5x)×(7y)2 } -(7y)3 ] + [(5x)3 +{3 ×(5x)2 ×(7y)} +{3 ×(5x)×(7y)2 } + (7y)3 ]
By using the formula,
(a + b)3 = a3 + 3a2 b + 3ab2 + b3 and (a – b)3 = a3 – 3a2 b + 3ab2 – b3
= [125x3 – {3 × 25x2 × 7y} + {3 × 5x × 49y2 } – 343y3 ] + [125x3 + {3 × 25x2 × 7y} + {3 × 5x × 49y2 } + 343y3 ]
= [125x3 – 525x2 y + 735xy2 – 343y3 ] + [125x3 + 525x2 y + 735xy2 + 343y3 ]
= 125x3 – 525x2 y + 735xy2 – 343y3 + 125x3 + 525x2 y + 735xy2 + 343y3
= 250x3 + 1470xy2
∴ (5x – 7y)3 + (5x + 7y)3 = 250x3 + 1470xy2
Practice set 5.4 PAGE NO: 28
1. Expand:
(1) (2p + q + 5)2
(2) (m + 2n + 3r)2
(3) (3x + 4y – 5p)2
(4) (7m – 3n – 4k)2
Solution:
(1) (2p + q + 5)2
Let us expand the given expression:
(2p + q + 5)2 = (2p)2 +(q)2 +(5)2 + [ 2 ×(2p)×(q)] + [ 2 ×(q)×(5)] + [ 2 ×(2p)×(5)]
By using the formula,
(a + b +c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac
Here, a = 2p, b = q, c = 5
Now, substitute the value we get
= 4p2 + q2 + 25 + [4pq] + [10q] + [20p]
= 4p2 + q2 + 25 + 4pq + 10q + 20p
∴ (2p + q + 5)2 = 4p2 + q2 + 25 + 4pq + 10q + 20p
(2) (m + 2n + 3r)2
Let us expand the given expression:
(m + 2n + 3r)2 = (m)2 +(2n)2 +(3r)2 + [2 ×(m)× (2n)] + [2 ×(2n)×(3r)] + [2 ×(m)×(3r)]
By using the formula,
(a + b +c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac
Here, a = m, b = 2n, c = 3r
Now, substitute the value we get
= m2 + 4n2 + 9r2 + [4mn] + [12nr] + [6mr]
= m2 + 4n2 + 9r2 + 4mn + 12nr + 6mr
∴ (m + 2n + 3r)2 = m2 + 4n2 + 9r2 + 4mn + 12nr + 6mr
(3) (3x + 4y – 5p)2
Let us expand the given expression:
(3x + 4y – 5p)2 = (3x)2 +(4y)2 +(-5p)2 + [2 ×(3x) ×(4y)] + [2 ×(4y)×(- 5p)] + [2 ×(3x)×(-5p)]
By using the formula,
(a + b +c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac
Here, a = 3x, b = 4y, c = 5p
Now, substitute the value we get
= 9x2 + 16y2 + 25p2 + [24xy] + [– 40yp] + [– 30xp]
= 9x2 + 16y2 + 25p2 + 24xy – 40yp – 30xp
∴ (3x + 4y – 5p)2 = 9x2 + 16y2 + 25p2 + 24xy – 40yp – 30xp
(4) (7m – 3n – 4k)2
Let us expand the given expression:
(7m – 3n – 4k)2 = (7m)2 +(- 3n)2 +(- 4k)2 + [2 ×(7m)×(-3n)] + [2 ×(-3n)×(-4k)] + [2 × (7m)×(-4k)]
By using the formula,
(a + b +c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac
Here, a = 7m, b = -3n, c = -4k
Now, substitute the value we get
= 49m2 + 9n2 + 16k2 + [– 42mn] + [24nk] + [– 56mk]
= 49m2 + 9n2 + 16k2 – 42mn + 24nk – 56mk
∴ (7m – 3n – 4k)2 = 49m2 + 9n2 + 16k2 – 42mn + 24nk – 56mk
2. Simplify:
(1) (x – 2y + 3)2 + (x + 2y – 3)2
(2) (3k – 4r – 2m)2 –(3k + 4r – 2m)2
(3) (7a – 6b + 5c)2 + (7a + 6b – 5c)2
Solution:
(1) (x – 2y + 3)2 + (x + 2y – 3)2
Let us expand the given expression:
(x – 2y + 3)2 + (x + 2y – 3)2 = [(x)2 +(-2y)2 + (3)2 +{2 ×(x)×(- 2y)} +{2 ×(- 2y)× (3)} + {2 ×(x)×(3)}] + [(x)2 + (2y)2 + (-3)2 + {2×(x)×(2y)} +{2×(2y)×(- 3)} +{2 ×(x)×(-3)}]
By using the formula,
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac
= [x2 + 4y2 + 9 + {– 4xy} + {– 12y} + {6x}] + [x2 + 4y2 + 9 + {4xy} + {– 12y} + {– 6x}]
= [x2 + 4y2 + 9 – 4xy – 12y + 6x] + [x2 + 4y2 + 9 + 4xy – 12y – 6x]
= x2 + 4y2 + 9 – 4xy – 12y + 6x + x2 + 4y2 + 9 + 4xy – 12y – 6x
= 2x2 + 8y2 + 18 – 24y
∴ (x – 2y + 3)2 + (x + 2y – 3)2 = 2x2 + 8y2 + 18 – 24y
(2) (3k – 4r – 2m)2 –(3k + 4r – 2m)2
Let us expand the given expression:
(3k – 4r – 2m)2 –(3k + 4r – 2m)2 = [(3k)2 +(- 4r)2 + (- 2m)2 +{2 ×(3k)×(- 4r)} + {2 ×(- 4r)× (-2m)} +{2 ×(3k)×(- 2m)}] –[(3k)2 + (4r)2 +(- 2m)2 +{2 ×(3k)×(4r)} +{2 × (4r)×(- 2m)} +{2 ×(3k)×(-2m)}]
By using the formula,
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac
= [9k2 + 16r2 + 4m2 + {– 24kr} + {16rm} + {– 12km}] –[9k2 + 16r2 + 4m2 + {24kr} + {– 16rm} + {– 12km}]
= [9k2 + 16r2 + 4m2 – 24kr + 16rm – 12km] –[9k2 + 16r2 + 4m2 + 24kr – 16rm – 12km]
= 9k2 + 16r2 + 4m2 – 24kr + 16rm – 12km – 9k2 – 16r2 – 4m2 – 24kr + 16rm + 12km
= –48kr + 32rm
= 32rm – 48kr
∴ (3k – 4r – 2m)2 –(3k + 4r – 2m)2 = 32rm – 48kr
(3) (7a – 6b + 5c)2 + (7a + 6b – 5c)2
Let us expand the given expression:
(7a – 6b + 5c)2 +(7a + 6b – 5c)2 = [(7a)2 +(- 6b)2 +(5c)2 + {2 ×(7a)×(- 6b)} + {2 ×(- 6b)× (5c)} +{2 ×(7a)×(5c)}] + [(7a)2 +(6b)2 +(- 5c)2 +{2 ×(7a)×(6b)} + {2 ×(6b)× (- 5c)} +{2 ×(7a)×(-5c)}]
By using the formula,
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac
= [49a2 + 36b2 + 25c2 + {– 84ab} + {– 60bc} + {70ac}] + [49a2 + 36b2 + 25c2 + {84ab} + {- 60bc} + {– 70ac}]
= [49a2 + 36b2 + 25c2 – 84ab – 60bc + 70ac] + [49a2 + 36b2 + 25c2 + 84ab – 60bc – 70ac]
= 49a2 + 36b2 + 25c2 – 84ab – 60bc + 70ac + 49a2 + 36b2 + 25c2 + 84ab – 60bc – 70ac
= 98a2 + 72b2 + 50c2 – 120bc
∴ (7a – 6b + 5c)2 + (7a + 6b – 5c)2 = 98a2 + 72b2 + 50c2 – 120bc
Our experts have simplified the difficult problems into simpler steps, which can be easily solved by students. These solutions will help you in obtaining knowledge and strong command over the subject. As the Chapter is about Expansion Formulae, regular revision of important concepts and formulas over time is the best way to strengthen your concepts.
Many such exercise problems are given in the MSBSHSE Solutions book, students can refer to them to yield good results in the examination.