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Given:

(x + 6) is a factor of f(x) = x3 + 3x2 + 4x + P

Concept:

If x + a is a factor of any function then it will make the value of the function zero when the value of x from x + a = 0 will apply to it.

Calculation:

Put x = - 6 in the f(x) = 0

⇒ (- 6)³ + 3(- 6)² + 4(- 6) + P = 0

⇒ - 216 + 108 - 24 + P = 0

⇒ P = 240 - 108

∴ P = 132

I. x² + 14x - 147 = 0

⇒  x² + 21x – 7x - 147 = 0

⇒ (x + 21)(x - 7) = 0

⇒ x = -21, 7

II. 6y² - y - 5 = 0

⇒ 6y² - 6y + 5y - 5 = 0

⇒ (6y + 5)(y - 1) = 0

⇒ y = -5/6, 1

Hence, x > y and x < y so the relationship between x and y cannot be established.

Given∶

I. x² - 13x + 40 = 0

⇒ x² - 8x - 5x + 40 = 0

⇒ x(x - 8) - 5 (x - 8) = 0

⇒ (x - 5) (x - 8) = 0

⇒ x = 5, 8

II. y² - 15y + 54 = 0

⇒ y² - 9y - 6y + 54 = 0

⇒ y(y - 9) - 6 (y - 9) = 0

⇒ (y - 6) (y - 9) = 0

⇒ y = 6, 9

Comparison between x and y (via Tabulation):

I. x² - 13x + 40 = 0

⇒ x² - 8x - 5x + 40 = 0

⇒ (x - 8)(x - 5) = 0

⇒ x = 8, 5

II. y² - 7y + 12 = 0

⇒ y² - 4y - 3y + 12 = 0

⇒ (y - 4) (y - 3)

⇒ y = 4, 3

Hence, x > y

I. x² + 13x + 42 = 0 

⇒ x² + 7x + 6x + 42 = 0 

⇒ (x + 6)(x + 7) = 0

⇒ x = -6, -7

II. y² – 2y - 63 = 0 

⇒ y² – 9y + 7y - 63 = 0 

⇒ (y – 9)(y + 7) = 0

⇒ y = 9, -7

∴ x = y or the relationship between x and y cannot be established.

Given:

Our given equation are 17x + 7y = 18 and 51x + 21y = p

Concept used:

Two linear equation a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 then,

Unique solution, then (a₁/a₂) ≠ (b₁/b₂)

Coincident, then (a₁/a₂) = (b₁/b₂) = (c₁/c₂)

No solution, then (a₁/a₂) = (b₁/b₂) ≠ (c₁/c₂)

It is important to note that the converse is also true.

Calculation:

For coincident, (a₁/a₂) = (b₁/b₂) = (c₁/c₂)

Comparing with the above equation then,

17/51 = 7/21 = 18/p

⇒ 1/3 = 18/p

⇒ p = 54

∴ The required value of p is 54

Given:

(602 - 542)

Formula used:

(a² - b²) = (a - b) × (a + b)

Calculations:

(602 - 542)

⇒ (60 + 54) × (60 - 54)

⇒ 114 × 6

⇒ 684

∴ The value of (602 - 542) is 684.

Given:

Number - (3451)51 × (531)43

Calculation:

As you can see the unit digit of both the numbers are 1.

Also, any power of 1 will be 1.

Hence, the unit digit of the product is 1.

∴ The unit digit is 1.

Calculation:

l. x2 – 2x – 15 = 0

⇒ x² – (5 – 3)x – 15 = 0

⇒ x² – 5x + 3x – 15 = 0

⇒ x(x – 5) + 3(x – 5) = 0

⇒ (x + 3) (x – 5) = 0

⇒ x = 5, -3

II. y² + 15y + 54 = 0

⇒ y² + (9 + 6)y + 54 = 0

⇒ y² + 9y + 6y + 54 = 0

⇒ y(y + 9) + 6(y + 9) = 0

⇒ (y + 6) (y + 9) = 0

⇒ y = -6, -9

∴ x > y

GIVEN:

Inequalities x² – 4x – 21 < 0 and y² – 18y + 77 ≤ 0

CONCEPT:

Inequality

CALCULATION:

x² – 4x – 21 < 0

⇒ (x – 7) (x + 3) < 0

⇒ x = (-3, 7)

y² – 18y + 77 ≤ 0

⇒ (y – 7) (y – 11) ≤ 0

⇒ y = [7, 11]

Hence, x ≤ y

From equation I) we get,

x² = 256

x = + 16, - 16

From equation II) we get,

y² + 18y + 17

⇒  y² + 17y + y + 17

⇒ y (y + 17) + 1(y + 17)

⇒ (y + 17) (y + 1)

∴  y = -17, - 1

Comparison between x and y (via Tabulation):

∴ x = y or the relationship between x and y cannot be established.

Given:

I. 2x² – 14x + 20 = 0

II. 8y² – 32y + 24 = 0

Concept used:

Using factorization (Middle term split) method.

Calculation:

2x² – 14x + 20 = 0

⇒ 2x² – 10x – 4x + 20 = 0

⇒ 2x(x – 5) – 4(x – 5) = 0

⇒ (x – 5)(2x – 4) = 0

⇒ (x – 5) = 0 or (2x – 4) = 0

⇒ x = 5 or x = 2

⇒ x = 2, 5

8y² – 32y + 24 = 0

⇒ 8y² – 24y – 8y + 24 = 0

⇒ 8y(y – 3) – 8(y – 3) = 0

⇒ (y – 3)(8y – 8) = 0

⇒ (y – 3) = 0 or (8y – 8) = 0

⇒ y = 3 or y = 1

⇒ y = 1, 3

∴ The relation cannot be determined.

Equation I:

2x2 + 6x – 80 = 0

⇒ 2x2 – 10x + 16x – 80   = 0

⇒ (2x + 16) (x – 5) = 0

⇒ x = - 8 or x = 5

Equation II:

3y2 + 12y + 9 = 0

⇒ 3y2 + 3y + 9y + 9 = 0

⇒ (3y + 9) (y + 1) = 0

⇒ y = - 3 or y = - 1

Comparing the values of x and y,

I. 23x³ = 621

⇒ x³ = 27

⇒ x = 3

II. 35y² = 315

⇒ y² = 9

⇒ y = +3 or -3

Comparison between x and y (via Tabulation):

∴ x ≥ y

Given:

\(x^4 + \dfrac{1}{x^4}=322, x \neq 0\)      ----(1)

Formula used:

(a – b)² = a² + b² – 2ab

(a + b)2 = a2 + b2 + 2ab

Calculation:

(x² + 1/x²)² = x⁴ + 1/x⁴ + 2

⇒ (x2 + 1/x2)2 = 322 + 2      ----(from eq (1))

⇒ (x2 + 1/x2)2 = 324

⇒ (x2 + 1/x2) = (324)^1/2

⇒ (x2 + 1/x2) = 18      ----(2)

Now, (x – 1/x)² = x2 + 1/x² – 2

⇒ (x – 1/x)2 = 18 – 2      ----(from eq (2))

⇒ (x – 1/x) = 16^1/2

⇒ (x – 1/x) = 4

∴ The value of (x – 1/x) is 4

Given:

The difference of square of two numbers is 45 and the differences of the numbers are 3

Formula used:

(a² – b²) = (a + b) × (a - b)

Calculation:

According to the question the difference of square of two numbers is 45

∴ (a² – b²) = 45

And the differences of the numbers are 3

∴ (a - b) = 3

Now, put the values in the given identity

∴ 45 = (a + b) × 3

⇒ (a + b) = 45/3 = 15

Hence, option (1) is correct

Given:

(a - b) = √8 and (a + b) = √32

Concept Used:

Basic concept of arithmetic and surds

Calculation:

It is given that (a - b) = √8

∴ √8 can be written as = 2√2

⇒ (a - b) = 2√2      ---(1)

And (a + b) = √32

∴ √32 can be written as = 4√2

⇒ (a + b) = 4√2     ---(2)

By equating (1) and (2), we get

a - b + a + b = 2√2 + 4√2

⇒ a = 3√2 and b = √2

Hence, option (3) is correct

Given:

The ratio of number when 5 subtracted to it and 2 added to it is 5 ∶ 7

Concept used:

Basic Concept of ratio and proportion

Calculation:

Le the ratio be x

∴ When 5 is subtracted to it the number is (x - 5) and when 2 added to it the number will be (x + 2)

Now, the ratio is 5 ∶ 7

∴ (x - 5)/(x + 2) = 5/7

⇒ 2X = 45

⇒ X = 22.5

Hence, option (2) is correct.

Given:

From Vadodara station, if we buy two tickets to station P and three tickets to station Q, the total cost is Rs. 77, but if we buy 3 tickets to station P and 5 tickets to station Q, the total cost is Rs. 124

Concept used:

Linear equation in two variables.

Calculation:

Two tickets to station P and three tickets to station Q

⇒ 2P + 3Q = 77 - (i)

Three tickets to station P and 5 tickets to station Q

⇒ 3P + 5Q = 124 - (ii)

Solve the above two equations

⇒ (2P + 3Q = 77) × 5

⇒ (3P + 5Q = 124) × 3

⇒ 10P + 15Q = 385 - (iii) 

⇒ 9P + 15Q = 372 - (iv)

Solving Equation three and four we get

⇒ P = 385 - 372 = 13

∴ The Fare from Vadodara to station P is Rs.13

Equation I.

(x – 13)2 = 0

⇒ x – 13 = 0

⇒ x = 13

Equation II.

y2 = 169

⇒ y = √169

⇒ y = 13 or -13

∴ x ≥ y

Given:

x² – 4√2x + 6 = 0

Concept used:

By using factorization method (Middle term split)

Calculation:

x² – 4√2x + 6 = 0

⇒ x² – 3√2x – √2x + 6 = 0

⇒ x(x – 3√2) – √2(x – 3√2) = 0

⇒ (x – 3√2)(x – √2) = 0

⇒ (x – 3√2) = 0 or (x – √2) = 0

⇒ x = 3√2 or x = √2

∴ The roots of equation is √2 and 3√2.    

Given:

a²x² – 3abx + 2b² = 0

Concept used:

By using factorization method (Middle term split)

Calculation:

a²x² – 3abx + 2b² = 0

⇒ a²x² – 2abx – abx + 2b² = 0

⇒ ax(ax – 2b) – b(ax – 2b) = 0

⇒ (ax – 2b)(ax – b) = 0

⇒ (ax – 2b) = 0 or (ax – b) = 0

⇒ x = 2b/a or x = b/a

∴ The roots of equation are 2b/a and b/a.

Given:

2x + 5y = 12

xy = 8

Formula used:

(a + b)² = a² + b² + 2ab

(a³ + b³) = (a + b) (a2 + b2 - ab)

Calculations:

2x + 5y = 12

Both side square,

(2x + 5y)² = (12)²

⇒ 4x² + 25y² + 20xy = 144

⇒ 4x2 + 25y2 + 20 × 8 = 144

⇒ 4x2 + 25y2 = -16      

Now,

8x3 + 125y3 = (2x + 5y) (4x² + 25y² - 10xy)

⇒ (12) × (-16 - 10 × 8)

⇒ -1152

∴ The value of 8x3 + 125y3 is -1152.

Given:

x = (5 + 33^1/2)/4

Calculation:

4x - 5 = 33^1/2

Squaring both sides we get,

16x² + 25 - 40x = 33

16x² -40x - 8 = 0

2x² - 5x - 1 = 0       ----(1)

To find:

The value of 8x³ + 6x² -69x + 20 = ?

Calculation:

Taking eq(1) and multiplying it by some factors to get the evaluating part. firstly we will multiply eq(1) by 4x in order to get the first term of evaluating part (8x³) then we will multiply the 13.

4x (2x² - 5x - 1) + 13 (2x² - 5x - 1) = 0

On solving and addition and subtraction of 20 above we get,

8x³ + 6x² -69x -13 + 20 - 20 =0

8x³ + 6x² -69x + 20 = 33

Calculation:

m² + 105 = n²

⇒ n² – m² = 105

⇒ (n – m)(n + m) = 105

n, m are positive integer

Factor of 105 = 1, 3, 5, 7, 15, 21, 35, 105

⇒ n + m > n – m

Condition,

n + m = 105 & n – m = 1

Solve m & n

⇒ n = 53, m = 52

n + m = 35 & n – m = 3

Solve m & n

⇒ n = 19, m = 16

n + m = 21 & n – m = 5

Solve m & n

⇒ n = 13, m = 8

n + m = 15 & n – m = 7

Solve m & n

⇒ n = 11, m = 4

∴ There are four pair.

Given:

a + b = 11 and ab = 15

Formula:

(a + b)² = a² + b² + 2ab

Calculation:

(a + b)2 = a2 + b2 + 2ab

⇒ 11² = a² + b² + 2 × 15

⇒ a² + b² = 121 - 30

∴ a² + b² = 91

I. 5x2 - 14x -3 = 0

⇒ 5x² -15x + x - 3 =0

⇒ 5x( x - 3) + 1(x -3) = 0 

⇒ (5x +1)(x -3) = 0

⇒ x = -1/5, 3

II. 9y2 + 16y - 4 = 0

⇒ 9y2 -18y + 2y - 4 =0

⇒ 9y( y - 2) + 2(y -2) = 0 

⇒ (9y + 2)(y - 2) = 0

⇒ y = -2/9, 2

∴ No relation can established between x,y

Given:

5x² – 6x – 5 = 0

Calculation:

5x² – 5 = 6x

⇒ 5(x² – 1) = 6x

⇒ x² – 1 = 6x/5

Dividing x on both sides

⇒ x – (1/x) = 6/5

⇒ {x – (1/x)}² = 36/25

∴ The value of {x – (1/x)}² is 36/25.

Given:

Total amount = Rs. 312

The number of one rupee notes = number of five rupee notes = number of twenty rupee notes

Calculations:

Let the number of one rupee notes, five rupee notes and twenty rupee notes be x, each.

Now, on calculating the sum of the value of these notes, we get:

(1 × x) + (5 × x) + (20 × x) = 312

⇒ x = 12

So, the total number of notes = 12 + 12 + 12 = 36

∴ The total number of notes that he has is 36.

Given:

CP of (4 eraser + 6 pencil + 7 notebook) = Rs. 265

CP of (2 eraser + 3 pencil + 3 notebook) = Rs. 125

Calculation:

∵ CP of (4 eraser + 6 pencil + 7 notebook) = Rs. 265      ------(1)

CP of (2 eraser + 3 pencil + 3 notebook) = Rs. 125      ------(2)

Multiply (1) with 2 and (2) with 3;

∴ CP of (8 eraser + 12 pencil + 14 notebook) = Rs. 530      ------(3)

CP of (6 eraser + 9 pencil + 9 notebook) = Rs. 375      ------(4)

Subtracting (4) from (3);

CP of (2 eraser + 3 pencil + 5 notebook) = Rs. 155

Calculation:

From I,

x² + 12x + 32 = 0

⇒ x2 + 8x + 4x +32 = 0

⇒ x(x+8) + 4(x+8) = 0

⇒ (x+4) (x+8) = 0

Taking, 

⇒ x + 4 = 0 or x + 8 = 0

⇒ x = –4 or –8

From II, 

y² + 6y + 8 = 0

⇒ y² + 4y + 2y + 8 = 0

⇒ y(y+4) + 2(y+4) = 0 

⇒ (y+2) (y+4) = 0

Taking,

⇒ y + 2 = 0 or y + 4 = 0

⇒ y = –2 or –4

Comparison between x and y (via Tabulation):

∴ Relation between x and y is, x ≤ y. 

Given:

(x² – x – 6)

Calculations:

Using the given data,

⇒ (x² – x – 6) = (x² – 3x + 2x – 6)

⇒ x(x – 3) + 2(x – 3)

⇒ (x + 2)(x – 3)

∴ The factors of (x² – x – 6) are (x + 2)(x – 3)

Given:

2x + 3y = 12      ----(i)

4x – 5y = 2      ----(ii)

Calculations:

Multiplying (i) by 2,

⇒ 4x + 6y = 24      ----(iii)

Subtracting (ii) from (iii),

⇒ 6y – (-5y) = 22

⇒ 6y + 5y = 22

⇒ 11y = 22

⇒ y = 2

Substituting the value of y in (i),

⇒ 2x + (3 × 2) = 12

⇒ 2x + 6 = 12

⇒ 2x = 6

⇒ x = 3

∴ The value of x and y are 3 and 2 respectively.

Given:

2 bats and 5 balls costs Rs. 3100

3 bats and 3 balls costs Rs. 4020

Calculations:

Let the price of a bat be x and the price of a ball be y

⇒ 2x + 5y = 3100      ----(i)

⇒ 3x + 3y = 4020      ----(ii)

Multiplying (i) by 3 and (ii) by 2,

⇒ 6x + 15y = 9300      ----(iii)

⇒ 6x + 6y = 8040      ----(iv)

Subtracting (iv) from (iii),

⇒ 15y – 6y = 1260

⇒ 9y = 1260

⇒ y = Rs. 140

Substituting the value of y in (i),

⇒ 2x + (5 × 140) = 3100

⇒ 2x + 700 = 3100

⇒ 2x = 2400

⇒ x = Rs. 1200

Cost of 1 bat and 1 ball,

⇒ x + y = 1200 + 140

⇒ Rs. 1340

∴ The cost of a bat and a ball together is Rs. 1340

Calculation:

Sum of roots = – b

4a + 3a = – b

⇒ 7a = – b

⇒ 49a² = b²

Product of roots = c

4a × 3a = c

⇒ 12a² = c

b2 + c = 49a² + 12a²

⇒ 61a²

⇒ 61 × 16 = 976

⇒ 61 × 9 = 549

∴ Possible value of b2 + c is 549

I. (x + 6) [8 – (1/x)] = 0

⇒ (x + 6) [(8x – 1)/x]

⇒ (x + 6) (8x – 1) = 0

⇒ x = – 6, 1/8

II. y² + √576 = √1089

⇒ y² + 24 = 33

⇒ y² = 9

⇒ y = 3, –3

∴ While comparing the values of x and y, both root values of y lie between the root values of x

Given

Teesta's salary is always a fixed portion of Torsa's salary

Tista's income was ₹ 1,625

Torsa's income was ₹ 2,125

Torsa receives ₹ 2,720

Concept used

Proportional concept

Calculation 

Ratio of income of Tista and Torsa = 2125 : 2720

⇒ 13 : 17

∴ 17 part = 2720

⇒ 1 part = 160

⇒ 13 part = 160 × 13

⇒ Rs 2080

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⇒ \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaWcaaWdaeaapeGaaiiOaiaadIhapaWaaWbaaSqabeaapeWaaSaa % a8aabaWdbiaaiodaa8aabaWdbiaaikdaaaaaaOGaaiiOaiabgEna0k % aacckadaGcaaWdaeaapeGaamiEaaWcbeaakiaacckacaGGGcGaeyOe % I0IaaGioaiaaigdacaGGGcaapaqaa8qacaGGGcGaeyOgIyTaamiEaa % aacqGH9aqpcaaIWaaaaa!4B95! \frac{{\;{x^{\frac{3}{2}}}\; \times \;√ x \;\; - 81\;}}{{\;\surd x}} = 0\)\(\frac{{\;{x^{\frac{3}{2}}}\; \times \;√ x \;\; - 81\;}}{{\;\surd x}} = 0\)

⇒ x^3/2 ×  x^1/2 – 81 = 0

⇒ x² = 81

⇒ x = ± 9

II. 20y² – 119y + 176 = 0

⇒ 20y² – 64y – 55y + 176 = 0

⇒ 4y(5y – 16) – 11(5y – 16) = 0

⇒ (5y – 16)(4y – 11) = 0

⇒ \(y = \;\frac{{16}}{5}\;,\;\frac{{11}}{4}\)

∴ x = y or the relationship between x and y cannot be established

We can not take the negative value of x as in the question there is 

√x is positive

x² is both positive and negative

Put a = 2 and b = -3 in a³ – b³

(2)³ – (-3)³

⇒  8 – (-27)

⇒  8 + 27

2^{2x + 2} = 2⁰        (a⁰ = 1)

⇒  2x + 2 = 0

⇒  2x = -2

Given:

4(3x – 2) = 2(3x + 8)

Concept used:

Using the concept of linear equation in 1 variable.

Calculation:

4(3x – 2) = 2(3x + 8)

⇒ 12x – 8 = 6x + 16

⇒ 6x = 24

⇒ x = 4

∴ The value of x is equal to 4.

Given:

w = –2, x = 3, y = 0, and z = –1/2

Calculation:

\(\ x\sqrt{(x+wz)} = \ 3\sqrt{(3+(-2)\times (-1/2))}\)

 \(\Rightarrow \ x√{(x+wz)} = 3√{4}\)

As, √4 = 2

\(\therefore \ x\sqrt{(x+wz)} = 6\)

Given:

a2 + b2 = 80

a – b = 4

Formula used:

(a – b)² = a² + b² – 2ab

Calculation:

a – b = 4

⇒ (a – b)² = 4²               [Squaring both sides]

⇒ a2 + b2 – 2ab = 16

⇒ 80 – 2ab = 16

⇒ -2ab = 16 – 80

⇒ -2ab = -64

⇒ ab = 32

∴ The value of ab is 32.

Solution:

Let the larger area be x Km² and smaller area be y Km².

⇒ x + y = 500      ----(1)

According to the question, the difference of the areas of two parts is 4/5 of the average of two areas.

⇒ (x – y) = 4/5[(x + y)/2]            

⇒ (x – y) = 2/5(x + y)      ----(2)

Put the value of  (x + y) from equation (1) in equation (2),

⇒ (x - y) = 2/5 × 500

⇒ x – y = 200      ----(3)

On solving equation (1) and (3) we get,

⇒ x = 350 Km² and y = 150 Km²        

Hence the area of smaller part is 150 Km².

Solution:

Let the number be xy = 10x + y

And reverse of a number is yx = 10y + x

According to the question, x + y = 6      ----(1)

Also if 9 is subtracted from 4 times the number, the digits will interchanged.

⇒ 4(10x + y) – 9 = 10y + x

⇒ 40x + 4y -9 = 10y + x

⇒ 39x – 6y = 9      ----(2)

On solving equation (1) and (2)

⇒ x = 1 and y = 5

Hence the number is 15.

Given:

x⁴ + (1/x⁴) = 47

Formula used:

(x + y)² = x² + y² + 2xy

(x – y)² = x² + y² – 2xy

(x – y)³ =  x³ – y³ – 3xy (x – y)

Calculation:

x⁴ + (1/x⁴) = 47

⇒ [x² + (1/x²)]² = x⁴ + (1/x⁴) + 2 × x² × (1/x²)

⇒  [x² + (1/x²)]² = 47 + 2

⇒ [x² + (1/x²)]² = 49

⇒ [x² + (1/x²)] = 7

⇒ [x – (1/x)]² = x² + (1/x)² – 2× x × (1/x)

⇒  [x – (1/x)]² = 7 – 2

⇒ [x – (1/x)]² = 5

⇒ x – (1/x) =√5

⇒ [x – (1/x)]³ =  x³ – (1/x)³ – 3 × x × (1/x) (x – 1/x)

⇒ x³ – (1/x)³ = [ x – (1/x)]³ + 3 (x – 1/x)

⇒ x³ – (1/x)³ = (√5) ³ + 3 × √5 

⇒ x³ – (1/x)³ = 5√5 + 3√5 

⇒ x³ – (1/x)³ = 8√5 

∴The value of x³ – (1/x)³ is 8√5 

Alternate method:

x⁴ + (1/x⁴) = 47 = P

x² + (1/x²) = √(P + 2)

⇒ x² + (1/x²) = √(47 + 2)

⇒ x² + (1/x²) = 7 = Q

⇒ x – (1/x) = √(Q – 2)

⇒ x – (1/x) = √(7 – 2)

⇒ x – (1/x) = √5⇒ [x – (1/x)]³ =  x³ – (1/x)³ – 3 × x × (1/x) (x – 1/x)

⇒ x³ – (1/x)³ = [ x – (1/x)]³ + 3 (x – 1/x)

⇒ x³ – (1/x)³ = (√5) ³ + 3 × √5 

⇒ x³ – (1/x)³ = 5√5 + 3√5 

⇒ x³ – (1/x)³ = 8√5 

∴ The value of x³ – (1/x)³ is 8√5