Chapter 5 - Expansion Formulae
Practice set 5.1
1. Expand.
(1) (a + 2) (a – 1)
Solution:
We know that,
(x + a)(x + b) = x² + (a + b)x + ab
∴ (a + 2) (a – 1) = a² + (2 – 1) a + 2 × (– 1)
∴ (a + 2) (a – 1) = a² + a – 2
Ans: (a + 2) (a – 1) = a² + a – 2
(2) (m – 4) (m + 6)
Solution:
We know that,
(x + a)(x + b) = x² + (a + b)x + ab
∴ (m – 4) (m + 6) = m² + (– 4 + 6) m + (– 4) × 6
∴ (m – 4) (m + 6) = m² + 2m – 24
Ans: (m – 4) (m + 6) = m² + 2m – 24
(3) (p + 8) (p – 3)
Solution:
We know that,
(x + a)(x + b) = x² + (a + b)x + ab
∴ (p + 8) (p – 3) = p² + (8 – 3) p + 8 × (– 3)
∴ (p + 8) (p – 3) = p² + 5p – 24
Ans: (p + 8) (p – 3) = p² + 5p – 24
(4) (13 + x) (13 – x)
Solution:
We know that,
(x + a)(x + b) = x² + (a + b)x + ab
∴ (13 + x) (13 – x) = (13)² + (x – x) 13 + x × (– x)
∴ (13 + x) (13 – x) = 169 + 0 × 13 – x²
∴ (13 + x) (13 – x) = 169 – x²
Ans: (13 + x) (13 – x) = 169 – x²
(5) (3x + 4y) (3x + 5y)
Solution:
We know that,
(x + a)(x + b) = x² + (a + b)x + ab
∴ (3x + 4y) (3x + 5y) = (3x)² + (4y + 5y) 3x + 4y × 5y
∴ (3x + 4y) (3x + 5y) = 9x² + 9y × 3x + 20y²
∴ (3x + 4y) (3x + 5y) = 9x² + 27xy + 20y²
Ans: (3x + 4y) (3x + 5y) = 9x² + 27xy + 20y²
(6) (9x – 5t) (9x + 3t)
Solution:
We know that,
(x + a)(x + b) = x² + (a + b)x + ab
∴ (9x – 5t) (9x + 3t) = (9x)² + [(– 5t) + 3t] 9x + (– 5t) × 3t
∴ (9x – 5t) (9x + 3t) = 81x² + (– 2t) × 9x – 15t²
∴ (9x – 5t) (9x + 3t) = 81x² – 18xt – 15t²
Ans: (9x – 5t) (9x + 3t) = 81x² – 18xt – 15t²
(7) \(\large(\)m + \(\large \frac {2}{3})\) \(\large(\)m – \(\large \frac {7}{3})\)
Solution:
We know that,
(x + a)(x + b) = x² + (a + b)x + ab
∴ \(\large(\)m + \(\large \frac {2}{3})\) \(\large(\)m – \(\large \frac {7}{3})\) = m² + \(\large (\frac {2}{3} – \frac {7}{3})\) m + \(\large (\frac {2}{3})\) × – \(\large \frac {2}{3}\)
∴ \(\large(\)m + \(\large \frac {2}{3})\) \(\large(\)m – \(\large \frac {7}{3})\) = m² – \(\large \frac {5}{3}\) m – \(\large \frac {14}{9}\)
Ans: \(\large(\)m + \(\large \frac {2}{3})\) \(\large(\)m – \(\large \frac {7}{3})\) = m² – \(\large \frac {5}{3}\) m – \(\large \frac {14}{9}\)
(8) \(\large(\)\(x\) + \(\large \frac {1}{x})\) \(\large(\)\(x\) – \(\large \frac {1}{x})\)
Solution:
We know that,
(x + a)(x + b) = x² + (a + b)x + ab
∴ \(\large(\)\(x\) + \(\large \frac {1}{x})\) \(\large(\)\(x\) – \(\large \frac {1}{x})\) = \(x\)² + \(\large (\frac {1}{x} – \frac {1}{x})\) \(x\) + \(\large (\frac {1}{x})\) × – \(\large \frac {1}{x}\)
∴ \(\large(\)\(x\) + \(\large \frac {1}{x})\) \(\large(\)\(x\) – \(\large \frac {1}{x})\) = \(x\)² – 0\(x\) – \(\large \frac {1}{x²}\)
∴ \(\large(\)\(x\) + \(\large \frac {1}{x})\) \(\large(\)\(x\) – \(\large \frac {1}{x})\) = \(x\)² – \(\large \frac {1}{x²}\)
Ans: \(\large(\)\(x\) + \(\large \frac {1}{x})\) \(\large(\)\(x\) – \(\large \frac {1}{x})\) = \(x\)² – \(\large \frac {1}{x²}\)
(9) \(\large (\frac {1}{x}\) + 4\(\large)\) \(\large (\frac {1}{y}\) – 4\(\large)\)
Solution:
We know that,
(x + a)(x + b) = x² + (a + b)x + ab
∴ \(\large (\frac {1}{x}\) + 4\(\large)\) \(\large (\frac {1}{y}\) – 4\(\large)\) = \(\large (\frac {1}{y})\)² + (4 – 9) \(\large (\frac {1}{y})\) + 4 × (– 9)
∴ \(\large (\frac {1}{x}\) + 4\(\large)\) \(\large (\frac {1}{y}\) – 4\(\large)\) = \(\large \frac {1}{y²})\) + – 9 \(\large (\frac {5}{y})\) – 36
Ans: \(\large (\frac {1}{x}\) + 4\(\large)\) \(\large (\frac {1}{y}\) – 4\(\large)\) = \(\large \frac {1}{y²})\) + – 9 \(\large (\frac {5}{y})\) – 36
Practice set 5.2
1. Expand.
(1) (k + 4)³
Solution:
We know that,
(a + b)³ = a³ + 3a²b + 3ab² + b³
Here a = k and b = 4
∴ (k + 4)³ = (k)³ + 3(k)² (4) + 3(k)(4)² + (4)³
∴ (k + 4)³ = k³ + 12k² + 3(k)(16) + 64
∴ (k + 4)³ = k³ + 12k² + 48k + 64
Ans: (k + 4)³ = k³ + 12k² + 48k + 64
(2) (7x + 8y)³
Solution:
We know that,
(a + b)³ = a³ + 3a²b + 3ab² + b³
Here a = 7x and b = 8y
∴ (7x + 8y)³ = (7x)³ + 3(7x)² (8y) + 3(7x) (8y)² + (8y)³
∴ (7x + 8y)³ = 343x³ + 3(49x²)(8y) + 3(7x)(64y²) + 512y³
∴ (7x + 8y)³ = 343x³ + 1176x²y + 1344xy² + 512y³
Ans: (7x + 8y)³ = 343x³ + 1176x²y + 1344xy² + 512y³
(3) (7 + m)³
Solution:
We know that,
(a + b)³ = a³ + 3a²b + 3ab² + b³
Here a = 7 and b = m
∴ (7 + m)³ = (7)³ + 3(7)²(m) + 3(7)(m)² + (m)³
∴ (7 + m)³ = 343 + 3(49)(m) + 3(7)(m²) + m³
∴ (7 + m)³ = 343 + 147m + 21m² + m³
Ans: (7 + m)³ = 343 + 147m + 21m² + m³
(4) (52)³
Solution:
(52)³ = (50 + 2)³
We know that,
(a + b)³ = a³ + 3a²b + 3ab² + b³
Here a = 50 and b = 2
∴ (52)³ = (50)³ + 3(50)² (2) + 3(50)(2)² + (2)³
∴ (52)³ = 125000 + 3(2500)(2) + 3(50)(4) + 8
∴ (52)³ = 125000 + 15000 + 600 + 8
∴ (52)³ = 140608
Ans: (52)³ = 140608
(5) (101)³
Solution:
(101)³ = (100 + 1)³
We know that,
(a + b)³ = a³ + 3a²b + 3ab² + b³
Here a = 100 and b = 1
(101)³ = (100)³ + 3(100)²(1) + 3(100)(1)² + (1)³
∴ (101)³ = 1000000 + 3(10000) + 3(100) (1) + 1
∴ (101)³ = 1000000 + 30000 + 300 + 1
∴ (101)³ = 1030301
Ans: (101)³ = 1030301
(6)\(\large(\)\(x\) + \(\large \frac {1}{x})\)³
Solution:
We know that,
(a + b)³ = a³ + 3a²b + 3ab² + b³
Here a = \(x\) and b = \(\large \frac {1}{x}\)
∴ \(\large(\)\(x\) + \(\large \frac {1}{x})\)³ = (\(x\))³ + 3(\(x\))² \(\large (\frac {1}{x})\) + 3(\(x\))\(\large (\frac {1}{x})\)² + \(\large (\frac {1}{x})\)³
∴ \(\large(\)\(x\) + \(\large \frac {1}{x})\)³ = \(x\)³ + 3\(x\) + 3\(x\) \(\large (\frac {1}{x²})\) + \(\large (\frac {1}{x³})\)
∴ \(\large(\)\(x\) + \(\large \frac {1}{x})\)³ = \(x\)³ + 3\(x\) + \(\large \frac {3}{x}\) + \(\large (\frac {1}{x³})\)
Ans: \(\large(\)\(x\) + \(\large \frac {1}{x})\)³ = \(x\)³ + 3\(x\) + \(\large \frac {3}{x}\) + \(\large (\frac {1}{x³})\)
(7) \(\large(\)2m + \(\large \frac {1}{5})\)³
Solution:
We know that,
(a + b)³ = a³ + 3a²b + 3ab² + b³
Here a = 2m and b = \(\large \frac {1}{5}\)
∴ \(\large(\)2m + \(\large \frac {1}{5})\)³ = (2m)³ + 3(2m)² \(\large (\frac {1}{5})\) + 3(2m)\(\large (\frac {1}{5})\)² + \(\large (\frac {1}{5})\)³
∴ \(\large(\)2m + \(\large \frac {1}{5})\)³ = 8m³ + 3(4m²) \(\large (\frac {1}{5})\) + 3(2m)\(\large (\frac {1}{25})\) + \(\large \frac {1}{125}\)³
∴ \(\large(\)2m + \(\large \frac {1}{5})\)³ = 8m³ + \(\large (\frac {12m²}{5})\) + \(\large (\frac {6m}{25})\) + \(\large \frac {1}{125}\)
Ans: \(\large(\)2m + \(\large \frac {1}{5})\)³ = 8m³ + \(\large (\frac {12m²}{5})\) + \(\large (\frac {6m}{25})\) + \(\large \frac {1}{125}\)
(8) \(\large (\frac {5x}{y}\) + \(\large \frac {y}{5x})\)³
Solution:
We know that,
(a + b)³ = a³ + 3a²b + 3ab² + b³
Here a = \(\large \frac {5x}{y}\) and b = \(\large \frac {y}{5x}\)
∴ \(\large (\frac {5x}{y}\) + \(\large \frac {y}{5x})\)³ = \(\large (\frac {5x}{y})\)³ + 3 \(\large (\frac {5x}{y})\)² \(\large (\frac {y}{5x})\) + 3 \(\large (\frac {5x}{y})\) \(\large (\frac {y}{5x})\)² + \(\large (\frac {y}{5x})\)³
∴ \(\large (\frac {5x}{y}\) + \(\large \frac {y}{5x})\)³ = \(\large \frac {125x³}{y³}\) + 3 \(\large (\frac {25x²}{y²})\) \(\large (\frac {y}{5x})\) + 3 \(\large (\frac {5x}{y})\) \(\large (\frac {y²}{25x²})\) + \(\large \frac {y³}{125x³}\)
∴ \(\large (\frac {5x}{y}\) + \(\large \frac {y}{5x})\)³ = \(\large \frac {125x³}{y³}\) + 3 \(\large (\frac {5x}{y})\) + 3 \(\large (\frac {y}{5x})\) + \(\large \frac {y³}{125x³}\)
∴ \(\large (\frac {5x}{y}\) + \(\large \frac {y}{5x})\)³ = \(\large \frac {125x³}{y³}\) + \(\large \frac {15x}{y}\) + \(\large \frac {3y}{5x}\) + \(\large \frac {y³}{125x³}\)
Ans: \(\large (\frac {5x}{y}\) + \(\large \frac {y}{5x})\)³ = \(\large \frac {125x³}{y³}\) + \(\large \frac {15x}{y}\) + \(\large \frac {3y}{5x}\) + \(\large \frac {y³}{125x³}\)
Practice set 5.3
1. Expand.
(1) (2m – 5)³
Solution:
We know that,
(a – b)³ = a³ – 3a²b + 3ab² – b³
Here, a = 2m and b = 5
∴ (2m – 5)³ = (2m)³ – 3(2m)² (5) + 3(2m) (5)² – (5)³
∴ (2m – 5)³ = 8m³ – 3(4m²)(5) + 3(2m)(25) – 125
∴ (2m – 5)³ = 8m³ – 60m² + 150m – 125
Ans: (2m – 5)³ = 8m³ – 60m² + 150m – 125
(2) (4 – p)³
Solution:
We know that,
(a – b)³ = a³ – 3a²b + 3ab² – b³
Here, a = 4 and b = p
∴ (4 – p)³ = (4)³ – 3(4)²(p) + 3(4)(p)² – (p)³
∴ (4 – p)³ = 64 – 3(16)(p) + 3(4)(p²) – p³
∴ (4 – p)³ = 64 – 48p + 12p² – p³
Ans: (4 – p)³ = 64 – 48p + 12p² – p³
(3) (7x – 9y)³
Solution:
We know that,
(a – b)³ = a³ – 3a²b + 3ab² – b³
Here, a = 7x and b = 9y
∴ (7x – 9y)³ = (7x)³ – 3(7x)² (9y) + 3 (7x)(9y)² – (9y)³
∴ (7x – 9y)³ = 343x³ – 3(49x²)(9y) + 3(7x)(81y²) – 729y³
∴ (7x – 9y)³ = 343x³ – 1323x²y + 1701xy² – 729y³
Ans: (7x – 9y)³ = 343x³ – 1323x²y + 1701xy² – 729y³
(4) (58)³
Solution:
(58)³ = (60 – 2)³
We know that,
(a – b)³ = a³ – 3a²b + 3ab² – b³
Here, a = 60 and b = 2
∴ (58)³ = (60)³ – 3(60)²(2) + 3(60)(2)² – (2)³
∴ (58)³ = 216000 – 3(3600)(2) + 3(60)(4) – 8
∴ (58)³ = 216000 – 21600 + 720 – 8
∴ (58)³ = 195112
Ans: (58)³ = 195112
(5) (198)³
Solution:
(198)³ = (200 – 2)³
We know that,
(a – b)³ = a³ – 3a²b + 3ab² – b³
Here, a = 200 and b = 2
∴ (198)³ = (200)³ – 3(200)²(2) + 3(200)(2)² – (2)³
∴ (198)³ = 8000000 – 3(40000)(2) + 3(200)(4) – 8
∴ (198)³ = 8000000 – 240000 + 2400 – 8
∴ (198)³ = 7762392
Ans: (198)³ = 7762392
(6) \(\large(\)2p – \(\large \frac {1}{2p})\)³
Solution:
We know that,
(a – b)³ = a³ – 3a²b + 3ab² – b³
Here, a = 2p and b = \(\large \frac {1}{2p}\)
∴ \(\large(\)2p – \(\large \frac {1}{2p})\)³ = (2p)³ – 3(2p)² \(\large (\frac {1}{2p})\) + 3(2p)\(\large (\frac {1}{2p})\)² – \(\large (\frac {1}{2p})\)³
∴ \(\large(\)2p – \(\large \frac {1}{p})\)³ = 8p³ – 3(2p) + 3\(\large (\frac {1}{2p})\) – \(\large (\frac {1}{8p³})\)
∴ \(\large(\)2p – \(\large \frac {1}{p})\)³ = 8p³ – 6p + \(\large (\frac {3}{2p})\) – \(\large (\frac {1}{8p³})\)
Ans: \(\large(\)2p – \(\large \frac {1}{p})\)³ = 8p³ – 6p + \(\large (\frac {3}{2p})\) – \(\large (\frac {1}{8p³})\)
(7) \(\large(\)1 – \(\large \frac {1}{a})\)³
Solution:
We know that,
(a – b)³ = a³ – 3a²b + 3ab² – b³
Here, a = 1 and b = \(\large \frac {1}{a}\)
∴ \(\large(\)1 – \(\large \frac {1}{a})\)³ = (1)³ – 3(1)² \(\large (\frac {1}{a})\) + 3(1)\(\large (\frac {1}{a})\)² – \(\large (\frac {1}{a})\)³
∴ \(\large(\)1 – \(\large \frac {1}{a})\)³ = 1 – 3\(\large (\frac {1}{a})\) + 3\(\large (\frac {1}{a²})\) – \(\large \frac {1}{a³}\)
∴ \(\large(\)1 – \(\large \frac {1}{a})\)³ = 1 – \(\large (\frac {3}{a})\) + \(\large (\frac {3}{a²})\) – \(\large \frac {1}{a³}\)
Ans: \(\large(\)1 – \(\large \frac {1}{a})\)³ = 1 – \(\large (\frac {3}{a})\) + \(\large (\frac {3}{a²})\) – \(\large \frac {1}{a³}\)
(8) \(\large (\frac {x}{3}\) – \(\large \frac {3}{x})\)³
Solution:
We know that,
(a – b)³ = a³ – 3a²b + 3ab² – b³
Here, a = \(\large \frac {x}{3}\) and b = \(\large \frac {3}{x}\)
∴ \(\large (\frac {x}{3}\) – \(\large \frac {3}{x})\)³ = \(\large (\frac {x}{3})\)³ – 3 \(\large (\frac {x}{3}\)² \(\large (\frac {3}{x}\) + 3 \(\large (\frac {x}{3}\) \(\large (\frac {3}{x}\)² – \(\large (\frac {3}{x}\)³
∴ \(\large (\frac {x}{3}\) – \(\large \frac {3}{x})\)³ = \(\large \frac {x³}{27}\) – 3 \(\large (\frac {x²}{9}\) \(\large (\frac {3}{x}\) + 3 \(\large (\frac {x}{3}\) \(\large (\frac {9}{x²}\) – \(\large (\frac {27}{x³}\)
∴ \(\large (\frac {x}{3}\) – \(\large \frac {3}{x})\)³ = \(\large \frac {x³}{27}\) – x + \(\large \frac {9}{x}\) – \(\large (\frac {27}{x³}\)
Ans: \(\large (\frac {x}{3}\) – \(\large \frac {3}{x})\)³ = \(\large \frac {x³}{27}\) – x + \(\large \frac {9}{x}\) – \(\large (\frac {27}{x³}\)
2. Simplify.
(1) (2a + b)³ – (2a – b)³
Solution:
We know that,
(a + b)³ = a³ + 3a²b + 3ab² + b³
(a – b)³ = a³ – 3a²b + 3ab² – b³
∴ (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]
∴ (2a + b)³ – (2a – b)³ = 8a³ + 12a²b + 6ab² + b³ – 8a³ + 12a²b – 6ab² + b³
∴ (2a + b)³ – (2a – b)³ = 8a³ – 8a³ + 12a²b + 12a²b + 6ab² – 6ab² + b³ + b³
∴ (2a + b)³ – (2a – b)³ = 24a²b + 2b³
Ans: (2a + b)³ – (2a – b)³ = 24a²b + 2b³
(2) (3r – 2k)³ + (3r + 2k)³
Solution:
We know that,
(a + b)³ = a³ + 3a²b + 3ab² + b³
(a – b)³ = a³ – 3a²b + 3ab² – b³
∴ (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]
∴ (3r – 2k)³ + (3r + 2k)³ = (27r³ – 54r²k + 36rk² – 8k³) + (27r³ + 54r²k + 36rk² + 8k³)
∴ (3r – 2k)³ + (3r + 2k)³ = 27r³ – 54r²k + 36rk² – 8k³ + 27r³ + 54r²k + 36rk² + 8k³
∴ (3r – 2k)³ + (3r + 2k)³ = 27r³ + 27r³ – 54r²k + 54r²k + 36rk² + 36rk² – 8k³ + 8k³
∴ (3r – 2k)³ + (3r + 2k)³ = 54r³ + 72rk²
Ans: (3r – 2k)³ + (3r + 2k)³ = 54r³ + 72rk²
(3) (4a – 3)³ – (4a + 3)³
Solution:
We know that,
(a + b)³ = a³ + 3a²b + 3ab² + b³
(a – b)³ = a³ – 3a²b + 3ab² – b³
∴ (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]
∴ (4a – 3)³ – (4a + 3)³ = (64a³ – 144a² + 108a – 27) – (64a³ + 144a² + 108a + 27)
∴ (4a – 3)³ – (4a + 3)³ = 64a³ – 144a² + 108a – 27 – 64a³ – 144a² – 108a – 27
∴ (4a – 3)³ – (4a + 3)³ = 64a³ – 64a³ – 144a² – 144a² + 108a – 108a – 27 – 27
∴ (4a – 3)³ – (4a + 3)³ = – 288a² – 54
Ans: (4a – 3)³ – (4a + 3)³ = – 288a² – 54
(4) (5x – 7y)³ + (5x + 7y)³
Solution:
We know that,
(a + b)³ = a³ + 3a²b + 3ab² + b³
(a – b)³ = a³ – 3a²b + 3ab² – b³
∴ (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]
∴ (5x – 7y)³ + (5x + 7y)³ = (125x³ – 525x²y + 735xy² – 343y³) + (125x³ + 525x²y + 735xy² + 343y³)
∴ (5x – 7y)³ + (5x + 7y)³ = 125x³ – 525x²y + 735xy² – 343y³ + 125x³ + 525x²y + 735xy² + 343y³
∴ (5x – 7y)³ + (5x + 7y)³ = 125x³ + 125x³ – 525x²y + 525x²y + 735xy² + 735xy² – 343y³ + 343y³
∴ (5x – 7y)³ + (5x + 7y)³ = 250x³ + 1470xy²
Ans: (5x – 7y)³ + (5x + 7y)³ = 250x³ + 1470xy²
Practice set 5.4
1. Expand.
(1) (2p + q + 5)²
Solution:
We know that,
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac
Here, a = 2p, b = q, c = 5,
∴ (2p + q + 5)² = (2p)² + (q)² + (5)² + 2(2p) (q) + 2(q) (5) + 2(2p) (5)
∴ (2p + q + 5)² = 4p² + q² + 25 + 4pq + 10q + 20p
Ans: (2p + q + 5)² = 4p² + q² + 25 + 4pq + 10q + 20p
(2) (m + 2n + 3r)²
Solution:
We know that,
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac
Here, a = m, b = 2n, c = 3r,
∴ (m + 2n + 3r)² = (m)² + (2n)² + (3r)² + 2(m) (2n) + 2(2n) (3r) + 2(m) (3r)
∴ (m + 2n + 3r)² = m² + 4n² + 9r² + 4mn + 12nr + 6mr
Ans: (m + 2n + 3r)² = m² + 4n² + 9r² + 4mn + 12nr + 6mr
(3) (3x + 4y – 5p)²
Solution:
We know that,
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac
Here, a = 3x, b = 4y, c = – 5p,
∴ (3x + 4y – 5p)² = (3x)² + (4y)² + (– 5p)² + 2(3x) (4y) + 2(4y) (– 5p) + 2(3x) (– 5p)
∴ (3x + 4y – 5p)² = 9x² + 16y² + 25p² + 24xy – 40py – 30px
Ans: (3x + 4y – 5p)² = 9x² + 16y² + 25p² + 24xy – 40py – 30px
(4) (7m – 3n – 4k)²
Solution:
We know that,
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac
Here, a = 7, b = – 3, c = – 4,
∴ (7m – 3n – 4k)² = (7m)² + (– 3n)² + (– 4k)² + 2(7m) (– 3n) + 2 (– 3n) (– 4k) + 2 (7m) (– 4k)
∴ (7m – 3n – 4k)² = 49m² + 9n² + 16k² – 42mn + 24nk – 56km
Ans: (7m – 3n – 4k)² = 49m² + 9n² + 16k² – 42mn + 24nk – 56km
2. Simplify.
(1) (x – 2y + 3)² + (x + 2y – 3)²
Solution:
We know that,
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac
Here, a = x, b = – 2y, c = 3,
∴ (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)]
∴ (x – 2y + 3)² + (x + 2y – 3)² = x² + 4y² + 9 – 4xy – 12y + 6x + x² + 4y² + 9 + 4xy – 12y – 6x
∴ (x – 2y + 3)² + (x + 2y – 3)² = x + x² + 4y² + 4y² + 9 + 9 – 4xy + 4xy – 12y – 12y + 6x – 6x
∴ (x – 2y + 3)² + (x + 2y – 3)² = 2x² + 8y² + 18 – 24y
Ans: (x – 2y + 3)² + (x + 2y – 3)² = 2x² + 8y² + 18 – 24y
(2) (3k – 4r – 2m)² – (3k + 4r – 2m)²
Solution:
We know that,
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac
Here, a = 3k, b = – 4r, c = – 2m,
∴ (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)]
∴ (3k – 4r – 2m)² – (3k + 4r – 2m)² = (9k² + 16r² + 4m² – 24kr + 16rm – 12km) – (9k² + 16r² + 4m² + 24kr – 16rm – 12km)
∴ (3k – 4r – 2m)² – (3k + 4r – 2m)² = 9k² + 16r² + 4m² – 24kr + 16rm – 12km – 9k² – 16r² – 4m² – 24kr + 16rm + 12km
∴ (3k – 4r – 2m)² – (3k + 4r – 2m)² = 32rm – 48kr
Ans: (3k – 4r – 2m)² – (3k + 4r – 2m)² = 32rm – 48kr
(3) (7a – 6b + 5c)² + (7a + 6b – 5c)²
Solution:
We know that,
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac
Here, a = 7, b = – 3, c = – 4,
∴ (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)]
∴ (7a – 6b + 5c)² + (7a + 6b – 5c)² = = 49a² + 36b² + 25c² – 84ab – 60bc + 70ac + 49a² + 36b² + 25c² + 84ab – 60bc – 70ac
∴ (7a – 6b + 5c)² + (7a + 6b – 5c)² = = 49a² + 49a² + 36b² + 36b² + 25c² + 25c² – 84ab + 84ab – 60bc – 60bc + 70ac – 70ac
∴ (7a – 6b + 5c)² + (7a + 6b – 5c)² = 98a² + 72b² + 50c² – 120bc
Ans: (7a – 6b + 5c)² + (7a + 6b – 5c)² = 98a² + 72b² + 50c² – 120bc