# MSBSHSE Solutions For Class 8 Maths Part 1 Chapter 5- Expansion Formulae

MSBSHSE Solutions For Class 8 Maths Part 1 Chapter 5 – Expansion Formulae are readily available here. This Chapter mainly deals with expansion of different forms of expressions. Mathematics is one of the scoring subjects where students secure maximum marks in the exam. When it comes to preparing for the exam, it is the toughest time where most students struggle to solve problems. So, here at BYJUâ€™S, our expert faculty team have developed Maharashtra State Board Class 8 Textbooks Part 1 solutions, which help students effortlessly prepare for their exams. All the solutions are well designed, keeping in mind the latest Maharashtra Board patterns and guidelines. Students can also learn easy tricks and short cut techniques when practised on a regular basis. The PDF of Maharashtra Board Solutions for Class 8 Maths Chapter 5 Expansion Formulae is available here, which students can download for free, from the links mentioned.

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Practice set 5.1 PAGE NO: 24

1. Expand:

Solution:

(1) (a + 2) (a â€“ 1)

Let us simplify the expression, we get

(a + 2) (a â€“ 1) = a2Â + [(2) + (-1)] a + [(2) Ã— (-1)]

By using the logic,

(x + p) (x + q) = x2Â + (p + q)x +(p Ã— q)

Here, x = a, p = 2, q = -1

Now, substitute the value we get

= a2Â + (2 â€“ 1)a + (-2)

= a2Â + 2a â€“ a â€“ 2

= a2Â + a â€“ 2

âˆ´ (a + 2) (a â€“ 1) = a2Â + a â€“ 2

(2) (m â€“ 4) (m + 6)

Let us simplify the expression, we get

(m â€“ 4) (m + 6) = m2Â + [(- 4) + (6)] m + [(- 4) Ã— (6)]

By using the logic,

(x + p) (x + q) = x2Â + (p + q)x + (p Ã— q)

Here, x = m, p = -4, q = 6

Now, substitute the value we get

= m2Â + (6 â€“ 4)m + (- 24)

= m2Â + 6m â€“ 4m â€“ 24

= m2Â + 2m â€“ 24

âˆ´ (m â€“ 4) (m + 6) = m2Â + 2m â€“ 24

(3) (p + 8) (p â€“ 3)

Let us simplify the expression, we get

(p + 8) (p â€“ 3) = p2Â + [(8) + (- 3)] p + [(8) Ã— (- 3)]

By using the logic,

(x + a) (x + b) = x2Â + (a + b)x +(a Ã— b)

Here, x = p, a = 8, b = -3

Now, substitute the value we get

= p2Â + (8 â€“ 3)p + (- 24)

= p2Â + 8p â€“ 3p â€“ 24

= p2Â + 5p â€“ 24

âˆ´ (p + 8) (p â€“ 3) = p2Â + 5p â€“ 24

(4) (13 + x) (13 â€“ x)

Let us simplify the expression, we get

(13 + x) (13 â€“ x) = (13)2Â â€“ (x)2

{We know that (a + b) (a â€“ b) = (a)2Â â€“ (b)2}

= 169 + 0(13) â€“ x2

= 169 â€“ x2

âˆ´ (13 + x) (13 â€“ x) = 169 â€“ x2

(5) (3x + 4y) (3x + 5y)

Let us simplify the expression, we get

(3x + 4y) (3x + 5y) = (3x)2Â + [(4y) + (5y)]Â 3x + [(4y) Ã— (5y)]

By using the logic,

(x + a) (x + b) = x2Â + (a + b)x + (a Ã— b)

Here, x = 3x, a = 4y, b = 5y

Now, substitute the value we get

= 9x2Â +Â [(9y) Ã— (3x)]Â + 20y2

= 9x2Â + 27xy + 20y2

âˆ´ (3x + 4y) (3x + 5y) = 9x2Â + 27xy + 20y2

(6) (9x â€“ 5t) (9x + 3t)

Let us simplify the expression, we get

(9x â€“ 5t) (9x + 3t) = (9x)2Â + [(-Â 5t) + (3t)]Â 9x + [(-Â 5t) Ã— (3t)]

By using the logic,

(x + a) (x + b) = x2Â + (a + b)x + (a Ã— b)

Here, x = 9x, a = -5t, b = 3t

Now, substitute the value we get

= 81x2Â +Â [(-Â 2t) Ã— (9x)]Â + (-Â 15t2)

= 81x2Â â€“ 18xt â€“ 15t2

âˆ´ (9x â€“ 5t) (9x + 3t) = 81x2Â â€“ 18xt â€“ 15t2

Practice set 5.2 PAGE NO: 25

1. Expand:

Solution:

(1) (k + 4)3

Let us simplify the expression, we get

(k + 4)3Â = (k)3Â + [Â 3 Ã—(k)2Â Ã— (4)] + [Â 3 Ã— (k) Ã— (4)2Â ] + (4)3

By using the formula,

(a + b)3Â = a3Â + 3a2b + 3ab2Â + b3

Here aÂ = k, bÂ = 4

Now, substitute the value we get

= k3Â + (3Â Ã—Â 4)k2Â + (3Â Ã—Â 16)k + 64

= k3Â + 12k2Â + 48k + 64

âˆ´ (k + 4)3Â = k3Â + 12k2Â + 48k + 64

(2) (7x + 8y)3

Let us simplify the expression, we get

(7x + 8y)3Â = (7x)3Â + [Â 3 Ã— (7x)2Â Ã— (8y)] + [Â 3 Ã— (7x) Ã— (8y)2Â ]Â + (8y)3

By using the formula,

(a + b)3Â = a3Â + 3a2b + 3ab2Â + b3

Here aÂ = 7x, bÂ = 8y

Now, substitute the value we get

= 343x3Â + (3Â Ã—Â 49Â Ã—Â 8)x2y + (3Â Ã—Â 7Â Ã—Â 64)xy2Â + 512y3

= 343x3Â + 1176x2y + 1344xy2Â + 512y3

âˆ´ (7x + 8y)3Â = 343x3Â + 1176x2y + 1344xy2Â + 512y3

(3) (7 + m)3

Let us simplify the expression, we get

(7 + m)3Â = (7)3Â + [Â 3 Ã— (7)2Â Ã— (m)] + [Â 3 Ã— (7) Ã— (m)2] + (m)3

By using the formula,

(a + b)3Â = a3Â + 3a2b + 3ab2Â + b3

Here aÂ = 7, bÂ = m

Now, substitute the value we get

= 343 + (3Â Ã—Â 49)m + (3Â Ã—Â 7)m2Â + m3

= 343 + 147m + 21m2Â + m3

âˆ´ (7 + m)3Â = 343 + 147m + 21m2Â + m3

(4) (52)3

Let us simplify the expression, we get

(52)3Â = (50 + 2)3

(50 + 2)3Â = (50)3Â + [3 Ã— (50)2Â Ã— (2)] + [Â 3 Ã— (50) Ã— (2)2] + (2)3

By using the formula,

(a + b)3Â = a3Â + 3a2b + 3ab2Â + b3

Here aÂ = 50, bÂ = 2

Now, substitute the value we get

= 125000 + (3Â Ã—Â 2500Â Ã—Â 2) + (3Â Ã—Â 50Â Ã—Â 4) + 8

= 125000 + 15000 + 600 + 8

= 140608

âˆ´ (52)3Â = (50 + 2)3 = 140608

(5) (101)3

Let us simplify the expression, we get

(101)3Â = (100 + 1)3

(100 + 1)3Â = (100)3Â + [Â 3 Ã—(100)2Â Ã—(1)] + [Â 3 Ã—(100)Ã—(1)2Â ] +(1)3

By using the formula,

(a + b)3Â = a3Â + 3a2b + 3ab2Â + b3

Here aÂ = 100, bÂ = 1

Now, substitute the value we get

= 1000000 + (3Â Ã—Â 10000Â Ã—Â 1) + (3Â Ã—Â 100Â Ã—Â 1) + 1

= 1000000 + 30000 + 300 + 1

= 1030301

âˆ´ (101)3Â = (100 + 1)3 = 1030301

Practice set 5.3 PAGE NO: 27

1. Expand:

Solution:

(1) (2m â€“ 5)3

Let us simplify the expression, we get

(2m â€“ 5)3Â = (2m)3Â â€“ [Â 3 Ã— (2m)2Â Ã— 5Â ] + [Â 3 Ã— (2m) Ã— (5)2] â€“ (5)3

By using the formula,

(a â€“ b)3Â = a3Â â€“ 3a2b + 3ab2Â â€“ b3

Here, aÂ = 2m, bÂ = -5

Now, substitute the value we get

= 8m3â€“ [3Â Ã—Â 4m2Â Ã—Â 5] + [ 3Â Ã—Â 2mÂ Ã—Â 25]Â â€“Â 125

= 8m3Â â€“ 60m2Â + 150m â€“ 125

âˆ´ (2m â€“ 5)3Â = 8m3Â â€“ 60m2Â + 150m â€“ 125

(2) (4 â€“ p)3

Let us simplify the expression, we get

(4 â€“ p)3Â = (4)3Â â€“ [Â 3 Ã— (4)2Â Ã— pÂ ] + [Â 3 Ã— (4) Ã— (p)2Â ] â€“ (p)3

By using the formula,

(a â€“ b)3Â = a3Â â€“ 3a2b + 3ab2Â â€“ b3

Here, aÂ = 4, bÂ = -p

Now, substitute the value we get

= 64Â â€“Â [3Â Ã—Â 6Â Ã—Â p ] + [ 3Â Ã—Â 4Â Ã—Â p2Â ]Â â€“Â p3

= 64 â€“ 48p + 12p2Â â€“ p3

âˆ´ (4 â€“ p)3Â = 64 â€“ 48p + 12p2Â â€“ p3

(3) (7x â€“ 9y)3

Let us simplify the expression, we get

(7x â€“ 9y)3Â = (7x)3Â â€“ [Â 3 Ã— (7x)2Â Ã— 9yÂ ] + [3 Ã— (7x) Ã— (9y)2Â ] â€“ (9y)3

By using the formula,

(a â€“ b)3Â = a3Â â€“ 3a2b + 3ab2Â â€“ b3

Here, aÂ = 7x, bÂ = -9y

Now, substitute the value we get

= 343x3Â â€“Â [3Â Ã—Â 49x2 Ã—Â 9y] + [3Â Ã—Â 7xÂ Ã—Â 81y2]Â â€“Â 729y3

= 343x3Â â€“ 1323x2y + 1701xy2Â â€“ 729y3

âˆ´ (7x â€“ 9y)3Â = 343x3Â â€“ 1323x2y + 1701xy2Â â€“ 729y3

(4) (58)3

Let us simplify the expression, we get

(58)3Â = (60 â€“ 2)3

(60 â€“ 2)3Â = (60)3Â â€“ [3 Ã— (60)2Â Ã— 2] + [3 Ã— (60) Ã— (2)2] â€“ (2)3

By using the formula,

(a â€“ b)3Â = a3Â â€“ 3a2b + 3ab2Â â€“ b3

Here, aÂ = 60, bÂ = -2

Now, substitute the value we get

= 216000Â â€“Â [3Â Ã—Â 3600Â Ã—Â 2] + [3Â Ã—Â 60Â Ã—Â 4]Â â€“Â 8

= 216000 â€“ 21600 + 720 â€“ 8

= 195112

âˆ´ (58)3Â = (60 â€“ 2)3 = 195112

(5) (198)3

Let us simplify the expression, we get

(198)3Â = (200 â€“ 2)3

(200 â€“ 2)3Â = (200)3Â â€“ [3 Ã— (200)2Â Ã— 2] + [3 Ã— (200) Ã— (2)2] â€“ (2)3

By using the formula,

(a â€“ b)3Â = a3Â â€“ 3a2b + 3ab2Â â€“ b3

Here, aÂ = 200, bÂ = -2

Now, substitute the value we get

= 8000000 â€“ 240000 + 2400 â€“ 8

= 7762392

âˆ´ (198)3Â = (200 â€“ 2)3 = 7762392

2. Simplify:

(1) (2a + b)3Â â€“ (2a â€“ b)3

(2) (3r â€“ 2k)3Â + (3r + 2k)3

(3) (4a â€“ 3)3Â â€“ (4a + 3)3

(4) (5x â€“ 7y)3Â + (5x + 7y)3

Solution:

(1) (2a + b)3Â â€“ (2a â€“ b)3

Let us expand the given expression:

(2a + b)3Â â€“(2a â€“ b)3Â = [(2a)3Â +{3 Ã—(2a)2Â Ã— bÂ } + {3 Ã—(2a)Ã—(b)2Â } +(b)3Â ] – [(2a)3Â -{3 Ã—Â (2a)2Â Ã— bÂ } +{3 Ã—(2a)Ã—(b)2} -(b)3]

By using the formula,

(a + b)3Â = a3Â + 3a2b + 3ab2Â + b3Â and (a â€“ b)3Â = a3Â â€“ 3a2b + 3ab2Â â€“ b3

= [8a3Â + {3Â Ã—Â 4a2Ã—Â b} + {3Â Ã—Â 2aÂ Ã—b}Â + b3] â€“ [8a3Â â€“Â {3Â Ã—Â 4a2Ã—Â b} + {3Â Ã—Â 2aÂ Ã—Â b2}Â â€“ b3]

= [8a3Â + 12a2b + 6ab2Â + b3] â€“ [8a3Â â€“ 12a2b + 6ab2Â â€“ b3]

= 8a3Â + 12a2b + 6ab2Â + b3Â â€“ 8a3Â + 12a2b â€“ 6ab2Â + b3

= 24a2b + 2b3

âˆ´ (2a + b)3Â â€“ (2a â€“ b)3 = 24a2b + 2b3

(2) (3r â€“ 2k)3Â + (3r + 2k)3

Let us expand the given expression:

(3r â€“ 2k)3Â + (3r + 2k)3Â = [(3r)3Â -{3 Ã—(3r)2Â Ã—(2k)} + {3 Ã—(3r)Ã—(2k)2Â } -(2k)3Â ] + [(3r)3Â +{3 Ã— (3r)2Â Ã—(2k)} + {3 Ã—(3r)Ã—(2k)2} +(2k)3Â ]

By using the formula,

(a + b)3Â = a3Â + 3a2b + 3ab2Â + b3Â and (a â€“ b)3Â = a3Â â€“ 3a2b + 3ab2Â â€“ b3

= [27r3Â â€“ {3 Ã—Â 9r2Â Ã— 2k} + {3 Ã—Â 3rÂ Ã—Â 4k2} â€“ 8k3] + [27r3Â + {3 Ã—Â 9r2Â Ã— 2k} + {3 Ã—Â 3rÂ Ã—(4k2)} +Â 8k3]

= [27r3Â â€“ 54r2k + 36rk2Â â€“ 8k3] + [27r3Â + 54r2k + 36rk2Â + 8k3]

= 27r3Â – 54r2k + 36rk2Â â€“ 8k3Â + 27r3Â + 54r2k + 36rk2Â + 8k3

= 54r3Â + 72rk2

âˆ´ (3r â€“ 2k)3Â + (3r + 2k)3 = 54r3Â + 72rk2

(3) (4a â€“ 3)3Â â€“ (4a + 3)3

Let us expand the given expression:

(4a â€“ 3)3Â â€“ (4a + 3)3Â = [(4a)3Â – {3 Ã—(4a)2Â Ã— 3Â } + {3 Ã—(4a)Ã—(3)2Â } – (3)3Â ] – [(4a)3Â +{3Â Ã— (4a)2Â Ã— 3} + {3 Ã—(4a)Ã—(3)2} + (3)3]

By using the formula,

(a + b)3Â = a3Â + 3a2b + 3ab2Â + b3Â and (a â€“ b)3Â = a3Â â€“ 3a2b + 3ab2Â â€“ b3

= [64a3 – {3 Ã—Â 16a2Â Ã— 3} + {3 Ã— 4a Ã—Â 9}Â –Â 27] â€“ [64a3Â + {3Â Ã—Â 16a2Â Ã— 3} + {3 Ã— 4a Ã—Â 9}Â +Â 27]

= [64a3Â – 144a2Â + 108a – 27] â€“ [64a3Â + 144a2Â + 108a + 27]

= 64a3Â – 144a2Â + 108a â€“ 27 â€“ 64a3Â – 144a2Â – 108a – 27

= – 288a2Â â€“ 54

âˆ´ (4a â€“ 3)3Â â€“ (4a + 3)3 = -288a2Â â€“ 54

(4) (5x â€“ 7y)3Â + (5x + 7y)3

Let us expand the given expression:

(5x â€“ 7y)3Â + (5x + 7y)3Â = [(5x)3Â -{3 Ã—(5x)2Â Ã—Â (7y)} +{3 Ã—(5x)Ã—(7y)2} -(7y)3] + [(5x)3Â +{3 Ã—(5x)2Â Ã—(7y)} +{3 Ã—(5x)Ã—(7y)2} + (7y)3]

By using the formula,

(a + b)3Â = a3Â + 3a2b + 3ab2Â + b3Â and (a â€“ b)3Â = a3Â â€“ 3a2b + 3ab2Â â€“ b3

= [125x3Â â€“ {3 Ã—Â 25x2Â Ã—Â 7y} + {3 Ã— 5x Ã—Â 49y2} â€“ 343y3] + [125x3Â + {3 Ã—Â 25x2Â Ã— 7y} + {3 Ã— 5x Ã—Â 49y2}Â +Â 343y3]

= [125x3Â â€“ 525x2y + 735xy2Â â€“ 343y3] + [125x3Â + 525x2y + 735xy2Â + 343y3]

= 125x3Â â€“ 525x2y + 735xy2Â â€“ 343y3Â + 125x3Â + 525x2y + 735xy2Â + 343y3

= 250x3Â + 1470xy2

âˆ´ (5x â€“ 7y)3Â + (5x + 7y)3 = 250x3Â + 1470xy2

Practice set 5.4 PAGE NO: 28

1. Expand:

(1) (2p + q + 5)2

(2) (m + 2n + 3r)2

(3) (3x + 4y â€“ 5p)2

(4) (7m â€“ 3n â€“ 4k)2

Solution:

(1) (2p + q + 5)2

Let us expand the given expression:

(2p + q + 5)2Â = (2p)2Â +(q)2Â +(5)2Â + [ 2 Ã—(2p)Ã—(q)] + [ 2 Ã—(q)Ã—(5)] + [ 2 Ã—(2p)Ã—(5)]

By using the formula,

(a + b +c)2Â = a2Â + b2Â + c2Â + 2ab + 2bc + 2ac

Here, a = 2p, b = q, c = 5

Now, substitute the value we get

= 4p2Â + q2Â + 25 + [4pq] + [10q] + [20p]

= 4p2Â + q2Â + 25 + 4pq + 10q + 20p

âˆ´ (2p + q + 5)2 = 4p2Â + q2Â + 25 + 4pq + 10q + 20p

(2) (m + 2n + 3r)2

Let us expand the given expression:

(m + 2n + 3r)2Â = (m)2Â +(2n)2Â +(3r)2Â + [2 Ã—(m)Ã— (2n)] + [2 Ã—(2n)Ã—(3r)] + [2 Ã—(m)Ã—(3r)]

By using the formula,

(a + b +c)2Â = a2Â + b2Â + c2Â + 2ab + 2bc + 2ac

Here, a = m, b = 2n, c = 3r

Now, substitute the value we get

= m2Â + 4n2Â + 9r2Â + [4mn] + [12nr] + [6mr]

= m2Â + 4n2Â + 9r2Â + 4mn + 12nr + 6mr

âˆ´ (m + 2n + 3r)2 = m2Â + 4n2Â + 9r2Â + 4mn + 12nr + 6mr

(3) (3x + 4y â€“ 5p)2

Let us expand the given expression:

(3x + 4y â€“ 5p)2Â = (3x)2Â +(4y)2Â +(-5p)2Â + [2 Ã—(3x) Ã—(4y)] + [2 Ã—(4y)Ã—(- 5p)] + [2 Ã—(3x)Ã—(-5p)]

By using the formula,

(a + b +c)2Â = a2Â + b2Â + c2Â + 2ab + 2bc + 2ac

Here, a = 3x, b = 4y, c = 5p

Now, substitute the value we get

= 9x2Â + 16y2Â + 25p2Â + [24xy] + [â€“ 40yp] + [â€“ 30xp]

= 9x2Â + 16y2Â + 25p2Â + 24xy â€“ 40yp â€“ 30xp

âˆ´ (3x + 4y â€“ 5p)2 = 9x2Â + 16y2Â + 25p2Â + 24xy â€“ 40yp â€“ 30xp

(4) (7m â€“ 3n â€“ 4k)2

Let us expand the given expression:

(7m â€“ 3n â€“ 4k)2Â = (7m)2Â +(- 3n)2Â +(- 4k)2Â + [2 Ã—(7m)Ã—(-3n)] + [2 Ã—(-3n)Ã—(-4k)] + [2 Ã— (7m)Ã—(-4k)]

By using the formula,

(a + b +c)2Â = a2Â + b2Â + c2Â + 2ab + 2bc + 2ac

Here, a = 7m, b = -3n, c = -4k

Now, substitute the value we get

= 49m2Â + 9n2Â + 16k2Â + [â€“ 42mn] + [24nk] + [â€“ 56mk]

= 49m2Â + 9n2Â + 16k2Â â€“ 42mn + 24nk â€“ 56mk

âˆ´ (7m â€“ 3n â€“ 4k)2 = 49m2Â + 9n2Â + 16k2Â â€“ 42mn + 24nk â€“ 56mk

2. Simplify:

(1) (x â€“ 2y + 3)2Â + (x + 2y â€“ 3)2

(2) (3k â€“ 4r â€“ 2m)2Â – (3k + 4r â€“ 2m)2

(3) (7a â€“ 6b + 5c)2Â + (7a + 6b â€“ 5c)2

Solution:

(1) (x â€“ 2y + 3)2Â + (x + 2y â€“ 3)2

Let us expand the given expression:

(x â€“ 2y + 3)2Â + (x + 2y â€“ 3)2Â = [(x)2Â +(-2y)2Â + (3)2Â +{2Â Ã—(x)Ã—(- 2y)} +{2Â Ã—(- 2y)Ã—Â (3)} + {2Â Ã—(x)Ã—(3)}] +Â [(x)2Â + (2y)2Â + (-3)2Â + {2Ã—(x)Ã—(2y)} +{2Ã—(2y)Ã—(- 3)} +{2Â Ã—(x)Ã—(-3)}]

By using the formula,

(a + b + c)2Â = a2Â + b2Â + c2Â + 2ab + 2bc + 2ac

=Â [x2Â + 4y2Â + 9Â + {â€“ 4xy} + {â€“ 12y}Â + {6x}] + [x2Â + 4y2Â + 9 + {4xy}Â + {â€“Â 12y} + {â€“ 6x}]

= [x2Â + 4y2Â + 9 â€“ 4xy â€“ 12y + 6x] + [x2Â + 4y2Â + 9 + 4xy â€“Â 12y â€“ 6x]

= x2Â + 4y2Â + 9 â€“ 4xy â€“ 12y + 6x + x2Â + 4y2Â + 9 + 4xy â€“ 12y â€“ 6x

= 2x2Â + 8y2Â + 18 â€“ 24y

âˆ´ (x â€“ 2y + 3)2Â + (x + 2y â€“ 3)2 = 2x2Â + 8y2Â + 18 â€“ 24y

(2) (3k â€“ 4r â€“ 2m)2Â – (3k + 4r â€“ 2m)2

Let us expand the given expression:

(3k â€“ 4r â€“ 2m)2Â – (3k + 4r â€“ 2m)2Â = [(3k)2Â +(- 4r)2Â +Â (- 2m)2Â +{2Â Ã—(3k)Ã—(- 4r)} + {2Â Ã—(- 4r)Ã—Â (-2m)} +{2Â Ã—(3k)Ã—(- 2m)}] – [(3k)2Â + (4r)2Â +(- 2m)2Â +{2Â Ã—(3k)Ã—(4r)} +{2Â Ã— (4r)Ã—(- 2m)} +{2Â Ã—(3k)Ã—(-2m)}]

By using the formula,

(a + b + c)2Â = a2Â + b2Â + c2Â + 2ab + 2bc + 2ac

=Â [9k2Â + 16r2Â + 4m2 + {â€“ 24kr}Â + {16rm}Â + {â€“ 12km}] – [9k2Â +Â 16r2Â + 4m2Â + {24kr}Â + {â€“ 16rm}Â + {â€“ 12km}]

= [9k2Â + 16r2Â + 4m2Â â€“ 24kr + 16rm â€“ 12km] – [9k2Â + 16r2Â + 4m2Â + 24kr â€“ 16rm â€“ 12km]

= 9k2Â + 16r2Â + 4m2Â â€“ 24kr + 16rm â€“ 12km â€“ 9k2Â â€“ 16r2Â â€“ 4m2Â â€“ 24kr + 16rm + 12km

= – 48kr + 32rm

= 32rm â€“ 48kr

âˆ´ (3k â€“ 4r â€“ 2m)2Â – (3k + 4r â€“ 2m)2 = 32rm â€“ 48kr

(3) (7a â€“ 6b + 5c)2Â + (7a + 6b â€“ 5c)2

Let us expand the given expression:

(7a â€“ 6b + 5c)2Â +(7aÂ +Â 6bÂ â€“Â 5c)2Â = [(7a)2Â +(- 6b)2Â +(5c)2Â + {2Â Ã—(7a)Ã—(- 6b)} + {2Â Ã—(- 6b)Ã—Â (5c)} +{2Â Ã—(7a)Ã—(5c)}] + [(7a)2Â +(6b)2Â +(- 5c)2Â +{2Â Ã—(7a)Ã—(6b)} + {2Â Ã—(6b)Ã—Â (- 5c)} +{2Â Ã—(7a)Ã—(-5c)}]

By using the formula,

(a + b + c)2Â = a2Â + b2Â + c2Â + 2ab + 2bc + 2ac

=Â [49a2Â + 36b2Â + 25c2 + {â€“ 84ab} + {â€“ 60bc}Â +Â {70ac}] + [49a2Â + 36b2Â + 25c2Â + {84ab}Â + {-Â 60bc} + {â€“ 70ac}]

= [49a2Â + 36b2Â + 25c2Â â€“ 84ab â€“ 60bc + 70ac] + [49a2Â + 36b2Â + 25c2Â + 84ab â€“ 60bc â€“ 70ac]

= 49a2Â + 36b2Â + 25c2Â â€“ 84ab â€“ 60bc + 70ac + 49a2Â + 36b2Â + 25c2Â + 84ab â€“ 60bc â€“ 70ac

= 98a2Â + 72b2Â + 50c2Â â€“ 120bc

âˆ´ (7a â€“ 6b + 5c)2Â + (7a + 6b â€“ 5c)2 = 98a2Â + 72b2Â + 50c2Â â€“ 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.

## Frequently Asked Questions on Maharashtra State Board Solutions for Class 8 Maths Chapter 5 Expansion Formulae

### How are these Maharashtra State Board Class 8 Maths Chapter 5Â  Expansion Formulae Solutions helpful?

Students are encouraged to practise the questions and then to refer back to the solutions as it could help them analyse their performance. This will also help them to rectify the mistakes early on and avoid making any during the board exams.