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(i) (x – y) (x + y) (x² + y²) (x⁴ + y⁴)

= (x² – y²) (x² + y²) (x⁴ + y⁴)

= [(x²)² – (y²)²] (x⁴ + y⁴)

= (x⁴ – y⁴) (x⁴ – y⁴)

= [(x⁴)² – (y⁴)²]

= x⁸ – y⁸

(ii) (2x – 1) (2x + 1) (4x² + 1) (16x⁴ + 1)

= [(2x)² – (1)²] (4x² + 1) (16x⁴ + 1)

= (4x² – 1) (4x² + 1) (16x⁴ + 1) 1

= [(4x²)² – (1)²] (16x⁴ + 1) 1

= (16x⁴ – 1) (16x⁴ + 1) 1

= [(16x⁴)² – (1)²] 1

= 256x⁸ – 1

 (m² – n²m)² + 2m³n²

= (m²)² – 2 (m²) (n²) (m) + (n²m)² + 2m³n²

= m⁴ – 2m³n² + (n²m)² + 2m³n²

= m⁴+ n⁴m² – 2m³n² + 2m³n²

= m⁴+ m²n⁴

Sum of 3l – 4m – 7n² and 2l + 5m – 4n²

= 3l – 4m – 7n² + 2l + 3m – 4n²

= 3l + 2l – 4m + 3m – 7n² – 4n²

= 5l – m – 11n² ……………………..equation (1)

Sum of 9l + 2m – 3n² and -3l + m + 4n²

= 9l + 2m – 3n² + (-3l + m + 4n²)

= 9l – 3l + 2m + m – 3n² + 4n²

= 6l + 3m + n² ……………………….equation (2)

Let us subtract equation (i) from (ii), we get

= 6l + 3m + n² – (5l – m – 11n²)

= 6l – 5l + 3m + m + n² + 11n²

= l + 4m + 12n²

(i) 2a³ (3a + 5b)

= 2a³ (3a + 5b)

= (2a³ × 3a) + (2a³ × 5b)

= 6a⁴ + 10a³b

(ii) -11a (3a + 2b)

= -11a (3a + 2b)

= (-11a × 3a) + (-11a × 2b)

= -33a² – 22ab

(iii) -5a (7a – 2b)

= -5a (7a – 2b)

= (-5a × 7a) – (-5a × 2b)

= -35a² + 10ab

(iv) -11y² (3y + 7)

= -11y² (3y + 7)

= (-11y² × 3y) + (-11y² × 7)

= -33y³ – 77y²

Let us simplify the given expression

= 3/5 × -15/4 × 7/9 × x² × x × x² × y × y² × y²

= -7/4 × x²⁺¹⁺² × y¹⁺²⁺²

= 7/4x⁵y⁵

Now let us substitute when, x = 2 and y = -1

For -7/4x⁵y⁵

= -7/4 × (2)⁵ (-1)⁵

= -7/4 × 32 × -1

= 56

(i) (2x + y) (2x + y)

= 2x(2x + y) + y(2x + y)

= 4x² + 2xy + 2xy + y²

= 4x² + 4xy + y²

(ii) (a + 2b) (a – 2b)

= a(a – 2b) + 2b (a – 2b)

= a² – 2ab + 2ab – 4b²

= a² – 4b²

(iii) (a² + bc) (a² – bc)

= a² (a² – bc) + bc (a² – bc)

= a⁴ – a²bc + bca² – b²c²

= a⁴ – b²c²

Given,

 4/3a (a² + b² – 3c²)

= 4/3a (a² + b² – 3c²)

= (4/3a × a²) + (4/3a × b²) – (4/3a × 3c²)

= 4/3a³ + 4/3ab² – 4ac²

Given,

(x³ + 1/x³) (x³ – 1/x³)

= x³ (x³ – 1/x³) + 1/x³ (x³ – 1/x³)

= x⁶ – 1 + 1 – 1/x⁶

= x⁶ – 1/x⁶

We know that x + 1/x = 9

So when squaring both sides, we get

(x + 1/x)² = (9)²

x² + 2 × x × 1/x + (1/x)² = 81

x² + 2 + 1/x² = 81

x² + 1/x² = 81 – 2

x² + 1/x² = 79

Now again when we square on both sides we get,

(x² + 1/x²)² = (79)²

x⁴ + 2 × x² × 1/x² + (1/x²)² = 6241

x⁴ + 2 + 1/x⁴ = 6241

x⁴ + 1/x⁴ = 6241- 2

x⁴ + 1/x⁴ = 6239

∴ x⁴ – 1/x⁴ = 6239

Given,

(x⁴ + 2/x²) (x⁴ – 2/x²)

= x⁴ (x⁴ – 2/x²) + 2/x² (x⁴ – 2/x²)

= x⁸ – 2x² + 2x² – 4/x⁴

= (x⁸ – 4/x⁴)

Let we solve,

= 24x² (1 – 2x)

= (24x²× 1) – (24x²× 2x)

= 24x² – 48x³

Now let us evaluate the expression when x = 3

= 24x² – 48x³

= 24 × (3)² – 48 × (3)³

= 24 × (9) – 48 × (27)

= 216 – 1296

= -1080

(i) (4x/5 – 3y/4) (4x/5 + 3y/4)

= 4x/5 (4x/5 + 3y/4) – 3y/4 (4x/5 + 3y/4)

= 16/25x² + 12/20yx – 12/20xy – 9y²/16

= 16/25x² – 9/16y²

(ii) (2x + 3/y) (2x – 3/y)

= 2x (2x – 3/y) + 3/y (2x – 3/y)

= 4x² – 6x/y + 6x/y – 9/y²

= 4x² – 9/y²

(iii) (2a³ + b³) (2a³ – b³)

= 2a³ (2a³ – b³) + b³ (2a³ – b³)

= 4a⁶ – 2a³b³ + 2a³b³ – b⁶

= 4a⁶ – b⁶

Given,

 994 × 1006

We can express 994 as 1000 – 6 and 1006 as 1000 + 6

= 994 × 1006 

= (1000 – 6) (1000 + 6)

= 1000 (1000 + 6) – 6 (1000 + 6)

= 1000000 + 6000 – 6000 – 36

= 999964

We know that x + 1/x = 12

So when squaring both sides, we get

(x + 1/x)² = (12)²

x² + 2 × x × 1/x + (1/x)² = 144

x² + 2 + 1/x² = 144

x² + 1/x² = 144 – 2

x² + 1/x² = 142

When subtracting 2 from both sides, we get

x² + 1/x² – 2 × x × 1/x = 142 – 2

(x – 1/x)² = 140

x – 1/x = √140

Given,

x + y

We know that

x² + y² = 29

x² + y² + 2xy – 2xy = 29

(x + y)² – 2 (2) = 29

(x + y)² = 29 + 4

x + y = ± √33

We know that

x² + y² = 29

Squaring both sides, we get

(x² + y²)² = (29)²

x⁴ + y⁴ + 2x²y² = 841

x⁴ + y⁴ + 2 (2)² = 841

x⁴ + y⁴ = 841 – 8

x⁴ + y⁴ = 833

Let us we solve,

= -3y (xy+y²)

= (-3y × xy) + (-3y × y²)

= -3xy² – 3y³

Now let us evaluate the expression when x = 4 and y = 5

= -3xy² – 3y³

= -3 × (4) × (5)² – 3 × (5)³

= -300 – 375

= -675

(i) 35 × 37

We can express 35 as 30 + 5 and 37 as 30 + 7

= 35 × 37 

= (30 + 5) (30 + 7)

= 30 (30 + 7) + 5 (30 + 7)

= 900 + 210 + 150 + 35

= 1295

(ii) 53 × 55

We can express 53 as 50 + 3 and 55 as 50 + 5

= 53 × 55 

= (50 + 3) (50 + 5)

= 50(50 + 5) + 3 (50 + 5)

= 2500 + 250 + 150 + 15

= 2915

(iii) 103 × 96

We can express 103 as 100 + 3 and 96 as 100 – 4

= 103 × 96 

= (100 + 3) (100 – 4)

= 100 (100 – 4) + 3 (100 – 4)

= 10000 – 400 + 300 – 12

= 10000 – 112

= 9888

We know that: (x – y)² = x² + y² – 2xy 

Replace y with \(\frac{1}{x}\), we get 

\((x - \frac{1}{x})^2\) = x² + \(\frac{1}{x^2}\) – 2 

Since \((x^2 + \frac{1}{x^2}) = 51\)

\((x - \frac{1}{x})^2\) = 51-2 = 49

\((x - \frac{1}{x})\) = ±7 

Now, Find  \(x^3 - \frac{1}{x^3}\)

We know that, x³ – y³ = (x – y)(x² + y² + xy)

Replace y with \(\frac{1}{x}\) , we get 

x³ – \(\frac{1}{x^3}\) = (x – \(\frac{1}{x}\))(x² + \(\frac{1}{x^2}\) + 1) 

Use (x – \(\frac{1}{x}\)) = 7 and (x² + \(\frac{1}{x^2}\)) = 51 

 x³ – \(\frac{1}{x^3}\) = 7 x 52

x³ – \(\frac{1}{x^3}\) = 364

We know that the given equations are

2x + 3y = 14… equation (i)

2x – 3y = 2… equation (ii)

Now, let us square both the equations and subtract equation (ii) from equation (i), we get,

(2x + 3y)² – (2x – 3y)² = (14)2 – (2)²

4x² + 9y² + 12xy – 4x² – 9y² + 12xy = 196 – 4

24xy = 192

xy = 8

∴ The value of xy is 8.

Given: 2x + 3y = 13, xy = 6 

Cubing 2x + 3y = 13 both sides, we get 

(2x + 3y)³ = (13)³ 

(2x)³ + (3y)³ + 3(2x )(3y) (2x + 3y) = 2197 

8x³ + 27y³ + 18xy(2x + 3y) = 2197 

8x³ + 27y³ + 18 x 6 x 13 = 2197 

8x³ + 27y³ + 1404 = 2197 

8x³ + 27y³ = 2197 – 1404 

8x³ + 27y³ = 793

(i) 102 × 106

We can express 102 as 100 + 2 and 106 as 100 + 6

= 102 × 106

 = (100 + 2) (100 + 6)

= 100 (100 + 6) + 2 (100 + 6)

= 10000 + 600 + 200 + 12

= 10812

(ii) 109 × 107

We can express 109 as 100 + 9 and 107 as 100 + 7

= 109 × 107 

= (100 + 9) (100 + 7)

= 100 (100 + 7) + 9 (100 + 7)

= 10000 + 700 + 900 + 63

= 11663

Given,

a – b = 4, ab = 21

Choose a – b = 4 

Cubing both sides, 

(a – b)³ = (4)³ 

a³ – b³ – 3ab (a – b) = 64 

a³ – b³ – 3 × 21 x 4 = 64 

a³ – b³ – 252 = 64 

a³ – b³ = 64 + 252 

a³ – b³ = 316

\(x + \frac{1}{x} = 3\)

Squaring both sides,

\((x + \frac{1}{x})^2 = 3^2\)

\(x^2 + \frac{1}{x^2} + 2 = 9\)

\(x^2 + \frac{1}{x^2} = 9 -2 \)

\(x^2 + \frac{1}{x^2} = 7\)

a^{\(\frac{1}{3}\)} + b^{\(\frac{1}{3}\)} + c^{\(\frac{1}{3}\)} = 0

⇒ a^{\(\frac{1}{3}\)} + b^{\(\frac{1}{3}\)} = c^{\(\frac{1}{3}\)} ---------(i)

⇒ (a^{\(\frac{1}{3}\)})³ + (b^{\(\frac{1}{3}\)})³ = (-c^{\(\frac{1}{3}\)})³

⇒ a + b + 3.a^{\(\frac{1}{3}\)}.b^{\(\frac{1}{3}\)}( a^{\(\frac{1}{3}\)} + b^{\(\frac{1}{3}\)}) = -c

⇒  a + b + 3.a^{\(\frac{1}{3}\)}.b^{\(\frac{1}{3}\)}(-c)^{\(\frac{1}{3}\)} = -c

⇒ a + b + c = 3.a^{\(\frac{1}{3}\)}.b^{\(\frac{1}{3}\)}.c^{\(\frac{1}{3}\)}

⇒  (a + b + c)³ = (3.a^{\(\frac{1}{3}\)}.b^{\(\frac{1}{3}\)}.c^{\(\frac{1}{3}\)})³

⇒ (a + b + c)³ = 27abc

a + b + c = 9, ab + bc + ca = 26 

Squaring, a + b + c = 9 both sides, we get 

(a + b + c)² = (9)² 

a² + b² + c² + 2(ab + bc + ca) = 81

a² + b² + c² + 2 x 26 = 81 

a² + b² + c² + 52 = 81 

a² + b² + c² = 29 

Now, a³ + b³ + c³ – 3abc = (a + b + c) [(a² + b² + c² – (ab + bc + ca)] 

= 9[29 – 26] 

= 9 x 3 

= 27 

⇒ a³ + b³ + c³ – 3abc = 27

Given,

a + b + c = 0 and a² + b² + c² = 16

Choose a + b + c = 0 

Squaring both sides,

(a + b + c)² = 0 

a² + b² + c² + 2(ab + bc + ca) 

= 0 

16 + 2(ab + bc + c) = 0 

2(ab + bc + ca) = -16 

ab + bc + ca = \(\frac{-16}{2}\)

ab + bc + ca = -8

Here, we will use (a+b)² = a² + b² + 2ab

Given: 9x² + 25y² = 181 and xy = - 6

We write, (3x + 5y)² = (3x)² + (5y)² + 2 x 3x x x 5y

= 181 + 30xy

= 181 + 30(- 6) = 1

(3x + 5y)² = 1

(3x + 5y) = √1 = ±1

(i) 153² − 147² = (153 + 147) (153 − 147)

= (300) (6) = 1800

(iv) 12.1² − 7.9² = (12.1 + 7.9) (12.1 − 7.9) = (20.0) (4.2) = 84

(i) (102)²

So, 

= (102)² 

= (100 + 2)²

= (100)² + 2 (100) (2) + 2²

= 10000 + 400 + 4

= 10404

(ii) (99)²

So, 

= (99)² 

= (100 – 1)²

= (100)² – 2 (100) (1) + 1²

= 10000 – 200 + 1

= 9801

Given,

(703)²

We can express 700 as 700 + 3

So, 

= (703)² 

= (700 + 3)²

= (700)² + 2 (700) (3) + 3²

= 490000 + 4200 + 9

= 494209

Given ,

1.8 × 2.2

We can express 1.8 as 2 – 0.2 and 2.2 as 2 + 0.2

Using formula (a – b) (a + b) = a² – b²

We get,

1.8 × 2.2 

= (2 – 0.2) ( 2 + 0.2)

= (2)² – (0.2)²

= 4 – 0.04

= 3.96

(i) (82)² – (18)²

 Using formula (a – b) (a + b) = a² – b²

We get,

(82)² – (18)² 

= (82 – 18) (82 + 18)

= 64 × 100

= 6400

(ii) (467)² – (33)²

 Using formula (a – b) (a + b) = a² – b²

We get,

(467)² – (33)² 

= (467 – 33) (467 + 33)

= (434) (500)

= 217000

(iii) (79)² – (69)²

Using formula (a – b) (a + b) = a² – b²

We get,

(79)² – (69)² 

= (79 + 69) (79 – 69)

= (148) (10)

= 1480

(i) 197 × 203

We can express 203 as 200 + 3 and 197 as 200 – 3

Using formula (a – b) (a + b) = a² – b²

We get,

197 × 203 

= (200 – 3) (200 + 3)

= (200)² – (3)²

= 40000 – 9

= 39991

(ii) 113 × 87

We can express 113 as 100 + 13 and 87 as 100 – 13

Using formula (a – b) (a + b) = a² – b²

We get,

113 × 87 

= (100 – 13) (100 + 13)

= (100)² – (13)²

= 10000 – 169

= 9831

(iii) 95 × 105

We can express 95 as 100 – 5 and 105 as 100 + 5

Using formula (a – b) (a + b) = a² – b²

We get,

95 × 105 

= (100 – 5) (100 + 5)

= (100)² – (5)²

= 10000 – 25

= 9975

 (i) (9a + 1/6)²

= (9a)² + 2 (9a) (1/6) + (1/6)²

= 81a² + 3a + 1/36

(ii) (x + x²/2)²

= (x)² + 2 (x) (x²/2) + (x²/2)²

= x² + x³ + 1/4x⁴

(iii) (x/4 – y/3)²

(x/4)² – 2 (x/4) (y/3) + (y/3)²

= 1/16x² – xy/6 + 1/9y²

(i) (x + 2)²

= x² + 2 (x) (2) + 2²

= x² + 4x + 4

(ii) (8a + 3b)²

= (8a)² + 2 (8a) (3b) + (3b)²

= 64a² + 48ab + 9b²

(iii) (2m + 1)²

= (2m)² + 2(2m) (1) + 12

= 4m² + 4m + 1

Now,

(x² – ay)²

= (x²)² – 2 (x²) (ay) + (ay)²

= x⁴ – 2x²ay + a²y²

(i) (3x – 1/3x)²

= (3x)² – 2 (3x) (1/3x) + (1/3x)²

= 9x² – 2 + 1/9x²

(ii) (x/y – y/x)²

= (x/y)² – 2 (x/y) (y/x) + (y/x)²

= x²/y² – 2 + y²/x²

(iii) (3a/2 – 5b/4)²

= (3a/2)² – 2 (3a/2) (5b/4) + (5b/4)²

= 9/4a² – 15/4ab + 25/16b²

(iv) (a²b – bc²)²

= (a²b)² – 2 (a²b) (bc²) + (bc²)²

= a⁴b² – 2a²b²c² + b²c⁴

(i) \(({2x-1\over x})^2\)

\(= ({2x-1\over x})^2\)

= (2x)² + (\(1\over x\))² – 2 (2x)(\(1\over x\))   [(a – b)² = a² + b² – 2ab]

= \(4x^2 + 1 \over x^2 - 4\)

(ii) (2x + y) (2x – y)  

(2x + y) (2x – y) [(a – b)(a + b) = a² – b²] 

= (2x )² – (y)² 

= 4x² – y²

(iii) (a²b – b²a)² 

(a²b – b²a)² [(a – b)² = a² + b² – 2ab]

= (a²b)² + (b²a)² – 2 (a²b)( b²a) 

= a⁴b² + b⁴a² – 2a³b³ 

(iv) (a – 0.1) (a + 0.1)  

(a – 0.1) (a + 0.1) [(a – b)(a + b) = a² – b² ]

= (a)² – (0.1)² 

= (a)² – 0.01 

(v) (1.5 x² – 0.3y²) (1.5 x² + 0.3y²)

(1.5 x² – 0.3y²) (1.5x² + 0.3y²) [(a – b)(a + b) = a² – b² ] 

= (1.5 x²)² – (0.3y²)² 

= 2.25 x⁴ – 0.09y⁴

(i) We will use the identity, (a - b)² = a² + b² – 2ab

(399)² = (400 - 1)²

= (400)² + (1)² - 2 x 400.1

= 16000 + 1 - 800

= 159201

(ii) We will use the identity, (a - b)² = a² + b² – 2ab

(0.98)² = (1 - 0.02)²

 = (1)² + (0.02)² - 2 x 0.02 x 1

= 1 + 0.004 - 0.04

= 0.9604

(iii) We will use the identity, (a - b)(a + b)= a² - b²

991 x 1009 = (1000 - 9)(1000 + 9)

= (1000)² - (9)²

= 1000000 - 81

= 999919

(iv) We will use the identity, (a - b) (a + b)= a² - b²

117 x 83 = (100 + 17)(100 - 17)

= (100)² + (17)²

= 10000 - 289

= 9711

Here, we will use (a - b)² = a² + b² - 2ab

(x - \(\frac{1}{x}\)) = (-1)

(x - \(\frac{1}{x}\))² = (-1)²

(x)² + (\(\frac{1}{x}\))² - 2 x x x \(\frac{1}{x}\) = 1

x² + \(\frac{1}{x^2}\) = 1 + 2

x² + \(\frac{1}{x^2}\) = 3

Here, we will use (a+b)² = a² + b² +2ab

(x + \(\frac{1}{x}\)) = (11)

(x + \(\frac{1}{x}\))² = (11)²

(x)² + (\(\frac{1}{x}\))² + 2 x x x \(\frac{1}{x}\) = 121

(x)² + (\(\frac{1}{x^2}\)) = 121 - 2

(x)² + (\(\frac{1}{x^2}\)) = 119

 (x – y + 3z) from (2z – x – 3y)

The difference of two like terms is a like term whose coefficient is the difference of the numerical coefficient of the two like terms.

We have,

= (2z – x – 3y) – (x – y + 3z)

Change the sign of each term of the expression to be subtracted and then add.

= 2z – x – 3y – x + y – 3z

= (2z – 3z) + (– x – x) + (- 3y + y)

= (2 – 3)z + (– 1 – 1)x + (- 3 +1)y

= -1z – 2x – 2y

a – b = 5, ab = 12 

Squaring, a – b = 5, both sides, 

(a – b)² = (5)² 

a² + b² – 2ab = 25 

a² + b² – 2 x 12 = 25 

a² + b² – 24 = 25 

a² + b² = 25 + 24

a² + b² = 49

a³ + b³ = (a + b)³ - 3ab (a + b)

= (8)³ - 3(15)(8) = 512 - 360 = 152

(i) (a + b + c)² + (a − b + c)² 

= (a² + b² + c² + 2ab+2bc+2ca) + (a² + (−b)² + c² −2ab−2bc+2ca) 

= 2a² + 2b² + 2c² + 4ca 

(ii) (a + b + c)² − (a − b + c)² 

= (a² + b² + c² + 2ab + 2bc + 2ca) − (a² + (−b)² + c² − 2ab − 2bc + 2ca) 

= a² + b² + c² + 2ab + 2bc + 2ca − a² − b² − c² + 2ab + 2bc − 2ca 

= 4ab + 4bc 

(iii) (a + b + c)² + (a – b + c)² + (a + b − c)² 

= a² + b² + c² + 2ab + 2bc + 2ca + (a² + b² + (c)² − 2bc − 2cb + 2ca) + (a² + b² + c² + 2ab − 2bc – 2ab) 

= 3a² + 3b² + 3c² + 2ab − 2bc + 2ca 

(iv) (2x + p − c)² − (2x − p + c)² 

= [4x² + p² + c² + 4xp − 2pc − 4xc] − [4x² + p² + c² − 4xp− 2pc + 4xc] 

= 4x² + p² + c² + 4xp − 2pc − 4cx − 4x² − p² − c² + 4xp + 2pc− 4cx

= 8xp − 8xc 

= 8(xp − xc)

(v) (x² + y² − z²)² − (x² − y² + z²)² 

= (x² + y² + (−z)²)² − (x² − y² + z²)² 

= [x⁴ + y⁴ + z⁴ + 2x²y² + 2y²z² + 2x²z² − [x⁴ + y⁴ + z⁴ − 2x²y² − 2y²z² + 2x²z²] 

= 4x²y² – 4z²x²

a + b = 7, ab = 12 

Squaring, a + b = 7, both sides, 

(a + b)² = (7)² 

a² + b² + 2ab = 49

a² + b² + 2 x 12 = 49 

a² + b² + 24 = 49 

a² + b² = 49 - 24

a² + b² = 25

Given: 3x – 2y = 11 and xy = 12 

Cubing 3x – 2y = 11 both sides, we get

(3x – 2y)³ = (11)³ 

(3x)³ – (2y)³ – 3(3x)(2y) (3x – 2y) =1331 

27x³ – 8y³ – 18xy(3x -2y) =1331 

27x³ – 8y³ – 18 x 12 x 11 = 1331 

27x³ – 8y³ – 2376 = 1331 

27x³ – 8y³ = 1331 + 2376

27x³ – 8y³ = 3707

27x³ + 8y³ = (3x)³ + (2y)³ = (2x + 2y)³ - 3. 3x . 2y (3x + 2y)

(3x - 2y)³ - 18xy (3x + 2y)

= (9)³ - 18(3) (9) = 729 - 486 = 243

Given that,

(p² + 16) (p² – 1/4)

= p² (p² – 1/4) + 16 (p² – 1/4)

= p⁴ – 1/4p² + 16p² – 4

= p⁴ + 63/4p² – 4