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Given:

P(x) = Dividend = 2x³ + 3x² – 5kx + 15

Divisor = x – k

Remainder = 2k³

Calculation:

As the remainder is 2k³ when divided by x – k

For x – k = 0 , x = k

P(k) = 2k³

2k³ + 3k² – 5k² + 15 = 2k³

⇒ –2k² + 15 = 0

⇒ 2k² = 15

⇒ k² = 15/2

⇒ k = ±√(15/2)

∴ Value of k is ±√(15/2)

Given:

Kx² – 2√5x + 4 = 0

Concept used:

If the roots of the equation ax² + bx + c = 0 are real and equal,

Then b² – 4ac = 0

Calculation:

Kx² – 2√5x + 4 = 0

If roots are real and equal, then

b² – 4ac = 0

⇒ (-2√5)² – 4 × k × 4 = 0

⇒ 20 – 16k = 0

⇒ 16k = 20

⇒ k = 5/4

∴ The value of k is 5/4.

Given:

x2 - 5x + 1 = 0

Formula used:

(a + b)3 = a3 + b3 + 3ab(a + b)

Calculations:

x2 - 5x + 1 = 0

⇒ x(x - 5 + 1/x) = 0

⇒ (x - 5 + 1/x) = 0

⇒ (x + 1/x) = 5

Taking cube both sides,

⇒ (x + 1/x)³ = (5)³

⇒ x³ + 1/x³ + 3(x)(1/x)(x + 1/x) = 125

⇒ x³ + 1/x³ + 3(5) = 125

⇒ x3 + 1/x3 + 15 = 125

⇒ x3 + 1/x3 = 125 - 15

⇒ x3 + 1/x3 = 110

(x⁴ + 1/x²)/(x² + 1)

Dividing the numerator and denominator by x,

(1/x)(x4 + 1/x2)/(x2 + 1)(1/x)

⇒ (x³ + 1/x³)/(x + 1/x)

⇒ 110/5

⇒ 22

∴ The value of \(\left(x^4 + \frac 1 {x^2}\right) \div \left(x^2 + 1\right)\) is 22.

Given :

The salary of 2 teachers and 4 peons is 28000 rupees 

The salary of 2 teachers and 7 peons is 34000 rupees 

Calculations :

Let the salary of the teacher be 'T'

Let the salary of the peon be 'P'

According to the question 

2T + 4P = 28000      ....(1)

2T + 7P = 34000      ....(2)

Solve equ (1) and (2) 

T = 10000 and P = 2000

∴ The salary of the teacher will be 10000 rupees

Given-

a2x + b2y + c2z = 9x + 121y + 81z

Concept Used-

Comparison of both sides of an equation.

Calculation-

a2x + b2y + c2z = 9x + 121y + 81z

Comparing both sides of the above equation we get,

a = 3, b = 11, c = 9

c + a - b = 9 + 3 - 11

∴ c + a - b = 1

Given-

x + y = 5

xy = 6

Formula Used-

(a + b)² = (a - b)² + 4ab

Calculation-

x + y = 5

⇒ (x + y)² = 25

⇒ (x - y)² + 4xy = 25

⇒ (x - y)² = 25 - 4 × 6 

⇒ (x - y)² = 1

⇒ x - y = 1    [∵ x > y]

∴ The value of x - y is 1

Given:

The total amount which spends by A = Rs. 56,000

Price of one laptop = Rs. 2,000

Price increase = 75% 

Concepts used:

Expenditure = Price × Quantity

Calculation:

The initial quantity of laptops purchased by the individual at price Rs.2000 per unit is -

Price × Quantity = Expenditure

⇒ Rs. 2,000 × Quantity = Rs. 56,000

⇒ Quantity = Rs. 56,000 ÷ Rs.2,000

⇒ 28 units of laptops

According to the question

The new price of a laptop = Old price × (1 + 75%)

⇒ Rs. 2,000 × (1 + 0.75) 

⇒ Rs. 2,000 × 1.75

⇒ Rs. 3,500

New Price × New Quantity = Expenditure

⇒ Rs.3500 × Quantity = Rs.56000

⇒ Quantity = Rs.56000 ÷ Rs.3500

⇒ 16 units of laptops

Fall in the number of laptops purchased = Old quantity – new quantity 

⇒ (28 – 16) units

⇒ 12 units of laptops

∴ Quantity purchased of laptops has to be decreased by 12 units when the price of laptops rise by 75%, keeping the expenditure constant.

Given:

The total cost of 2 Kurtas and 5 shirts = Rs. 234

The cost of kurta is more than cost of shirt = Rs. 12

Calculation:

Let assume that cost of a shirt is Rs. x

So, cost of Kurta = Rs. x + 12

⇒ Cost of 5 shirts = 5 × x = Rs. 5x

⇒ Cost of 2 kurtas = 2 × (x + 12) = Rs. 2x + 24

According to the question

⇒ 5x + 2x + 24 = 234

⇒ 7x = 234 -24

⇒ 7x = 210

⇒ x = 210/7

⇒ x = 30

⇒ Cost of 1 Shirt = Rs. 30

⇒ Cost of 1 Kurta = Rs.30 + Rs. 12 = Rs. 42

∴ Cost of 3 shirts = Rs. 30 × 3 = Rs. 90

The correct option is 1 i.e. Rs. 90

I. x² - 7x + 12 = 0

⇒ x² - 4x - 3x + 12 = 0

⇒ (x - 4)(x - 3) = 0

⇒ x = 4, 3

II. y² - 5y + 6 = 0

⇒ y² - 3y - 2y + 6 = 0

⇒ (y - 3)(y - 2) = 0

⇒ y = 3, 2

Hence, x ≥ y

I. x2 - 7x + 12 = 0

⇒ x² - 4x - 3x + 12 = 0

⇒ x(x - 4) - 3( x - 4) = 0

⇒ (x - 3)(x - 4) = 0

⇒ x = 3, 4

y2 + 8x + 12 = 0 

⇒ y2 + 6y + 2y + 12 = 0

⇒ y(y + 6) + 2(y + 6) = 0

⇒ (y + 6)(y + 2) = 0

⇒ y = -6, -2

∴ x > y

GIVEN:

x = 9 – 4√5

CONCEPT USED:

a² - b² = (a + b) × (a - b)

CALCULATION:

x = 9 – 4√5

So,

1/x = 1/(9 – 4√5) = (9 + 4√5)/(9 – 4√5) × (9 + 4√5)

⇒ 1/x = (9 + 4√5)/(81 - 80) = 9 + 4√5

1/x = 9 + 4√5

⇒ x + 1/x = (9 – 4√5) + (9 + 4√5) = 18

We know that, if x + 1/x = a, then √x + 1/√x = √(a + 2)

Now,

√x + 1/√x = √(18 + 2) = √20

∴ 2√5

Given:

The cost of 2 tables and 3 chairs is Rs. 500

Concept Used:

If we have a linear equation in two variable and the required answer will be the multiple of the value of equation and a constant term then we don't need any other linear equation to find the answer

Calculation:

Let the cost of a table and a chair be Rs. x and Rs. y respectively

Now, according to question

2x + 3y = 500 ----(i)

and we have to find the value of 

8x + 12y

Now, if we do (i) × 4, we get 

4 × (2x + 3y) = 4 × 500

⇒ 8x + 12y = 2000

Hence, the price of 8 tables and 12 chairs is Rs. 2000

Calculation:

l. x² – 3x – 4 = 0

⇒ x² – 4x + x – 4 = 0

⇒ x(x – 4) + 1(x – 4) = 0

⇒ (x – 4) (x + 1) = 0

⇒ x = 4, -1.

ll. y² – 5y – 6 = 0

⇒ y² – 6y + y – 6 = 0

⇒ y(y – 6) + 1(y – 6) = 0

⇒ (y + 1) (y – 6) = 0

⇒ y = 6, -1.

Comparison between x and y (via Tabulation):

∴ The relation between x and y can not be established.

Calculation:

x2 – 9x + 14 = 0

⇒ x² - 7x - 2x + 14 = 0

⇒ x(x - 7) - 2(x - 7) = 0

⇒ (x - 2) (x - 7)

⇒ x = 2, 7.

y2 + y – 42 = 0

⇒ y² + 7y - 6y - 42 = 0

⇒ y(y + 7) - 6(y + 7) = 0

⇒ (y - 6) (y + 7) = 0

⇒ y = 6, -7.

Comparison between x and y (via Tabulation):

∴ The relation between x and y can not be determined.

Given:

x² + 3x – 28  = 0

⇒ x² + 7x – 4x – 28 = 0

⇒ x(x + 7) – 4 (x + 7) = 0

⇒ (x – 4) (x + 7) = 0

⇒ x = 4, -7.

y² – 7y + 10 = 0

⇒ y² – 5y – 2y + 10 = 0

⇒ y(y – 5) – 2(y – 5) = 0

⇒ (y – 5) (y – 2) = 0

⇒ y = 5, 2.

Comparison between x and y (via Tabulation):

∴ The relation between x and y can not be determine.

Calculation:

l. x² – 50x + 225 = 0

⇒ x² – 45x – 5x + 225 = 0                                                       

⇒ x(x - 45) – 5(x - 45) = 0

⇒ (x - 5) (x - 45) = 0

⇒ x = 5, 45

ll. y² + 32y – 105 = 0

⇒ y² + 35y – 3y – 105 = 0

⇒ y(y + 35) – 3(y + 35) = 0

⇒ (y - 3) (y + 35)

⇒ y = 3, -35

Comparison between x and y (via Tabulation):

∴ The result is x > y

Given:

a3 + 4a2 + 16a = 192

Calculations:

a3 + 4a2 + 16a = 192

When a = 4

L.H.S = 64 + 64 + 64 = 192

L.H.S. = R.H.S.

So, a3 + 8/a = 4³ + 8/4 = 64 + 2 = 66

∴ The value of a3 + 8/a is 66.

Given:

x3 + y3 + z3 = 17, x + y + z = 2 and xyz = 1

Formula used:

x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – zx)

(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

Calculations:

x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)

⇒ 17 – 3 × 1 = 2 × (x2 + y2 + z2 – xy – yz – zx)

⇒ 7 = x2 + y2 + z2 – xy – yz – zx

⇒ x2 + y2 + z2 = xy + yz + zx + 7

(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

⇒ (2)² = xy + yz + zx + 7 + 2xy + 2yz + 2zx

⇒ 3(xy + yz + zx) = -3

⇒ xy + yz + zx = -1

⇒ 2(xy + yz + zx) = -2

∴ The value of 2(xy + yz + zx) is -2.

Given:

[a + 1/(a + 7)] = (-5)

Concept:

If the value of the equation x + (1/x) is given as 2, then the value of x = 1.

Calculation:

∵ [a + 1/(a + 7)] = (-5)

Adding (7) on both sides;

⇒ [(a + 7) + 1/(a + 7)] = (-5) + 7

⇒ [(a + 7) + 1/(a + 7)] = 2

Now, this is in the form ‘x + (1/x) = 2’

∴ a + 7 = 1

⇒ [(a + 7)¹¹ + 1/(a + 7)¹²] = (1)¹¹ + 1/(1)¹²

= 1 + 1

= 2

Given:

3√3x³ + 2√2y³ – 8z³ + 6√6xyz = (Ax + By – Cz)(3x² + 2y² + 4z² – √2Axy + 2Byz + 2√3Zx)

Concept:

Compare the unknown values of the given equation with the actual values of the equation after expansion.

Formula used:

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

Calculation:

∵ 3√3x³ + 2√2y³ – 8z³ + 6√6xyz = (√3x)³ + (√2y)³ + (–2z)³ – 3(√3x)(√2y)( –2z)

∴ (√3x)³ + (√2y)³ + (-2z)³ – 3(√3x)(√2y)(-2z) = (√3x + √2y – 2z)[3x² + 2y² + 4z² – (√3x)(√2y) – (√2y)( –2z) – (√3x)(–2z)]

⇒ (√3x)³ + (√2y)³ + (-2z)³ – 3(√3x)(√2y)(-2z) = (√3x + √2y – 2z)[3x² + 2y² + 4z² – (√3x)(√2y) + (√2y)(2z) + (√3x)(2z)]

∴ If we compare the values of the equations;

We get

A = √3

B = √2

C = 2

∴ A² + B² – C² = (√3)² + (√2)² – (2)²

= 3 + 2 – 4

= 5 – 4

= 1

I.2y2 - 13y - 34 =0

⇒ 2y² + 4y - 17y - 34 = 0

⇒ 2y(y + 2) - 17 ( y + 2) = 0

⇒ (2y - 17) ( y + 2) = 0

⇒ y = 17/2, -2

II. 7x2 + 25x + 12 =0

⇒ 7x² + 21x + 4x + 12 = 0

⇒ 7x ( x  + 3) + 4 ( x+ 3) = 0 

⇒ (7x + 4) (x + 3) = 0

⇒ x = -4/3, -3

∴ x < y

I.16x2 -1 =0

⇒ 16x² = 1

⇒ (x) = + 1/4 , - 1/4

II. y2 = 1/4

⇒ y² = (1/2)²

⇒ y =  + 1/2 , - 1/2

∴ No relation can be established

Calculations:

From I,

x² – 10x + 21 = 0

⇒ x² – 7x – 3x + 21 = 0

⇒ x(x – 7) – 3(x – 7) = 0

⇒ (x – 7)(x – 3) = 0

Taking,

⇒ x – 7 = 0 or x – 3 = 0

⇒ x = 7 or x = 3

From II,

y² + 6y – 27 = 0

⇒ y² + 9y – 3y  – 27 = 0

⇒ y(y + 9) – 3(y + 9) = 0

⇒ (y + 9)(y – 3) = 0

Taking,

⇒ y + 9 = 0 or y – 3 = 0

⇒ y = –9 or y = 3

Comparison between x and y (via Tabulation):

∴ x ≥ y

I.x2 + 10x -24 = 0

⇒ x2 +12x -2x - 24 =0

⇒ x(x + 6) -2(x + 6) = 0

⇒ (x + 6) (x - 2) = 0

⇒ x = -6, 2 

II. 3y2 - 12y - 15 = 0

⇒ 3y2 - 15y + 3y - 15 =0

⇒ 3y(y - 5) + 3(y - 3) = 0

⇒ (3y + 5) (3y + 3) =0

⇒ y = 5/3, -1

Comparison between x and y (via Tabulation):

∴ There is no relation between x and y

x = y or relation between x and y can not be determined.

 Calculation:

I: x² + 10x – 299 = 0

⇒ x² + 23x – 13x – 299 = 0

⇒ x (x + 23) – 13(x + 23) = 0

⇒ (x + 23) × (x – 13) = 0

⇒ x = 13, (-23)

II: y² + 22y + 117 = 0

⇒ y² + 9y + 13y + 117 = 0

⇒ y × (y + 9) + 13 × (y + 9) = 0

⇒ (y + 9) × (y + 13) = 0

⇒ y = (-13), (-9)

Comparison between x and y (via Tabulation):

∴ From the table, we can conclude that x = y or relation cannot be established.

Calculation:

l. x² + 10x – 75 = 0

⇒ x² + (15 – 5)x – 75 = 0

⇒ x² + 15x – 5x – 75 = 0

⇒ x(x + 15) – 5(x + 15) = 0

⇒ (x + 15) (x – 5) = 0

⇒ x = -15, 5

II. 2y² – y – 136 = 0

⇒ 2y² – (17 – 16)y – 136 = 0

⇒ 2y² – 17y + 16y – 136 = 0

⇒ y(2y – 17) + 8(2y – 17) = 0

⇒ (2y – 17) (y + 8) = 0

⇒ y = 17/2, -8

∴ No relation in x and y or x = y

I. 6x² – 19x + 14 = 0

⇒ 6x² – 12x – 7x + 14 = 0

⇒ 6x(x – 2) – 7(x – 2) = 0

⇒ (x – 2)(6x – 7) = 0

⇒ x = 2, 7/6

II. y² + 14y + 48 = 0

⇒ y² + 6y + 8y + 48 = 0

⇒ y(y + 6) + 8(y + 6) = 0

⇒ (y + 6)(y + 8) = 0

⇒ y = – 6, – 8

∴ The relation between x and y is, x > y

From equation I) we get,

3x² + 19x + 20 = 0

⇒  3x² + 15x+ 4x + 20 = 0

⇒ 3x (x+ 5) + 4(x+5) + 20 = 0

⇒ (x+5) (3x + 4) = 0

∴ x = -5, - 4/3

From equation II) we get,

6y² + 19y + 15 = 0

⇒ 6y² + 10y + 9y + 15 = 0

⇒ 2y (3y + 5) + 3(3y + 5) = 0

⇒ (2y + 3) (3y + 5) = 0

∴  y = -3/2, -5/3

Comparison between x and y (via Tabulation):

∴ The relationship between x and y cannot be established.

Given:

Our given equation is 3x2 + 30x + 7

Concept used:

For maximum and minimum value, we can put

dy/dx = 0

And if d2y/dx2 > 0 then the value of x obtained from dy/dx will give minimum value when substituted in equation

And if d2y/dx2 < 0 then the value of x obtained from dy/dx will give maximum value when substituted in equation

Calculation:

Let y = 3x2 + 30x + 7

⇒ dy/dx = 6x + 30 = 0

⇒ x = -5

Now,  d2y/dx2 = 6 > 0,

y will give minimum value at x = -5

ymin = 3(-5)2 + 30 × (-5) + 7 = -68

∴ The minimum value of given expression is -68

Given:

The given quadratic equation = x² - 15x + 56

Calculation:

x² - 15x + 56 = 0

⇒ x² - 8x – 7x + 56 = 0

⇒ x(x - 8) - 7(x - 8) = 0

⇒ (x - 8)(x - 7) = 0

⇒ x = 7 and 8

So, α and β are 8 and 7 respectively.

GIVEN:

‘α’ and ‘β’ are the roots of expression x² – 12x + 27 = 0

FORMULA USED:

(α – β)2 = (α + β)2 – 4αβ

(α2 – β2) = (α – β)(α + β)

CALCULATION:

Since ‘α’ and ‘β’ are the roots of expression x² – 12x + 27 = 0.

α + β = -b/a = 12      ---- (1)

αβ = c/a = 27      ---- (2)

From (1) and (2):

(α – β)² = (α + β)² – 4αβ

⇒ (α – β)² = 144 – 108 = 36

⇒ α – β = 6

Now,

1/β² – 1/α²

⇒ (α² – β²)/α²β²

⇒ [(α – β)(α + β)]/α²β²

⇒ [(6 × 12)/27²]

⇒ 8/81

∴  The value of 1/β2 – 1/α2 is 8/81.

Given:

The given quadratic equation is 7x² + 4x – 1 = 0

Concept Used:

Sum of roots (α + β) = -b/a                                                  

And product of roots (α × β) = c/a

Calculation:

By comparing the given quadratic equation 7x² + 4x - 1 with standard equation ax² + bx + c we can find the value of coefficient a, b and c

∴ a = 7, b = 4 and c = -1

Sum of roots (α + β) = -b/a = - (4)/7 = -4/7

And, product of roots (α × β) = c/a = -1/7

We know that,

α² + β² = (α + β) ² - 2(α × β)

⇒ α² + β² = (-4/7) ² - 2(-1/7) = 30/49

Given:

Equation = -5a² + 8ab – 3a²  and 14a² – 11ab

Sum = -5a² + 8ab – 3a² + 14a² – 11ab

⇒ 6a² – 3ab

∴ The coefficient of a² is 6

Given:

Our given quadratic equation is ax2 + bx + c = 0  

Concept used:

General quadratic equation in terms of roots is x2 – (α + β)x + αβ = 0

Formula used:

a2 + b2 = (a + b)2 – 2ab

Calculation:

By comparing given quadratic equation with general quadratic equation, we get

(α + β) = –b/a and αβ = c/a

Now, 1/α2 + 1/β2

⇒ (α2 + β2)/(α2β2)

⇒ ((–b/a)2 – 2(c/a))/(c/a)2

⇒ (b2 – 2ca)/c2

∴ The required value of given expression is (b2 – 2ca)/c2

Given:

The equation 12x2 – ax + 7 = ax2 + 9x + 3

Formula used:

To have only one solution D = 0 

D = b² – 4ac

Calculation:

12x2 – ax + 7 = ax2 + 9x + 3

⇒ x²(12 – a) – x(9 + a) + 4 = 0

According to the question:

⇒ (9 + a)² – 4(12 – a)(4) = 0

⇒ 81 + a² + 18a – 192 + 16a = 0

⇒ a2 + 34a – 111 = 0

⇒ a2 + 37a – 3a – 111 = 0

⇒ a(a + 37) – 3(a + 37) = 0

⇒ (a – 3)(a + 37) = 0

If, (a – 3) = 0

⇒ a = 3

Either, (a + 37) = 

⇒ a = – 37 

⇒ a = – 37 is not positive.

∴ The positive internal solution of 'a' is 3.

I. 4x² – 72x + 224 = 0

⇒ 4x² – 56x – 16x + 224 = 0

⇒ 4x(x – 14) – 16(x – 14) = 0

⇒ (x – 14)(4x – 16) = 0

⇒ x = 14 or 4

II. y² – 19y + 60 = 0

⇒ y² – 15y – 4y + 60 = 0

⇒ y(y – 15) – 4(y – 15) = 0

⇒ (y – 15)(y – 4) = 0

⇒ y = 15 or 4

Comparison between x and y (via Tabulation):

∴ x = y or the relationship cannot be established.

From equation I) we get,

4x² + 13x + 9 = 0

⇒  4x² + 4x +9x + 9

⇒ 4x (x +1) +9(x + 1)

⇒ (x +1)(4x + 9)

x = -1, -9/4

From equation II) we get,

4y² + 20y + 25 =0

⇒ 4y² +10y +10Y +25

⇒ 2y(2y + 5) + 5(2y + 5)

⇒ (2y + 5) (2y + 5)

∴  y = -5/2, -5/2

Comparison between x and y (via Tabulation):

∴ x > y

Given:

Roots of quadratic equation are -1/3 and 1 

Calculation:

x = 1

m(1)² – n(1) – 1 = 0

⇒ m – n -1 = 0

⇒ m – n = 1                                                  …………………….. (1)

x = -1/3

m(-1/3)² – n(-1/3) – 1 = 0

⇒ m/9 + n/3 = 1

⇒ m + 3n = 9                                              ………………………. (2)

From equation (1) and (2)

-4n = -8

⇒ n = 2

Now,

⇒ m – 2 = 1

⇒ m = 3

∴ m = 3 and n = 2 

The correct option is 4 i.e. m = 3 and n = 2 

Given∶

I. 4x² - 7x + 3 = 0,

⇒ 4x² - 4x - 3x + 3 = 0,

⇒ 4x (x - 1) -3 (x - 1) = 0,

⇒ (4x - 3) (x - 1) = 0,

⇒ x = 3 / 4, 1

II. 16y² - 9 = 0,

⇒ 16y² = 9,

⇒ y² = 9 / 16,

⇒ y = ± 3 / 4

Given:

If a + b + c = 6, a2 + b2 + c2 = 30 and a3 + b3 + c3 = 165

Concept used:

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

a³ + b³ + c³ - 3abc = (a + b + c) (a2 + b2 + c² - ab - bc - ca)

Calculation:

⇒ (a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca)

⇒ 6² = 30 + 2 (ab + bc + ca)

⇒ (ab + bc + ca) = 6/2 = 3

⇒ a3 + b3 + c3 - 3abc = (a + b + c) (a2 + b2 + c2 - ab - bc - ca)

⇒ 165 - 3abc = 6 × (30 - 3)

⇒ 165 - 3abc = 6 × 27

⇒ 3abc = 165 - 162

⇒ abc = 1

∴ 4abc = 4 × 1 = 4

Given - 

x2 + 4y2 = 17 and xy = 2, where x > 0, y > 0

Formula used - 

a³ + b³ = (a + b) (a² - ab + b²)

(a + b)² = a² + b² + 2ab

Solution - 

x2 + 4y2 = 17 and xy = 2, where x > 0, y > 0

⇒ xy = 2 then, 4xy = 8

⇒ (x + 2y)² = x² + 4y² + 4xy = 17 + 8 = 25

⇒ (x + 2y) = 5

⇒ x3 + 8y³ = (x + 2y) (x² + 4y² - 2xy)

⇒ x3 + 8y³ = (5) × (17 - 4) = 65

∴ x3 + 8y³ = 65.

Short trick - 

⇒ put x = 1, y = 2

⇒ it satisfy both the given expression.

⇒x3 + 8y³ = 1³ + 8 × 2³ = 65