# Test: Algebraic Identities Cubic Type

## 20 Questions MCQ Test Mathematics (Maths) Class 9 | Test: Algebraic Identities Cubic Type

Description
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QUESTION: 1

### Expansion and simplification of 8(3h - 4) + 5(h - 2) gives

Solution:

Now, 8 (3h - 4) + 5 (h - 2)

= 24h - 32 + 5h - 10, the expansion

= 24h + 5h - 32 - 10

= 29h - 42, the simplification

QUESTION: 2

### Evaluate (11)3​

Solution:

11³= (10+1)³=1000+1+30(11)=1001+330=1331

QUESTION: 3

### Factorise:(3x- 5y)3+ (5y – 2z)3 + (2z – 3x)3

Solution:

We don't have to cube them all to factorise. Now, One thing should be remembered that, in time of this type of factorisation we have to first add the polynomials to check if the sum is zero or not.If it is zero, then the factorisation will be in form of 3(abc). Now, lets sum them, (3x-5y) +(5y-2z) +(2z-3x).The answer is zero. So put the expression in form of 3(abc). So the factorisation of the expression is 3(3x-5y)(5y-2z)(2z-3x).

QUESTION: 4

Evaluate: 53 – 23 – 33

Solution:

53 - 23 - 33​​​​​​​
= 125 - 8 - 27
= 90

QUESTION: 5

Factorise : 8a3+ b3 + 12a2b + 6a b2

Solution:

8a3 + b3+ 12a2b + 6ab2
8a3 + b3+ 12a2b + 6ab2
= (2a)3 + (b)3 + 3(2a)(b) (2a + b)
= (2a + b)3 | Using Identity VI
= (2a + b)(2a + b)(2a + b)

QUESTION: 6

Evaluate (x+y+z)2

Solution:
QUESTION: 7

Factorise : 8a3 + b3 + 12a2 b + 6ab2

Solution:

8a3 + b3+ 12a2b + 6ab2
8a3 + b3+ 12a2b + 6ab2
= (2a)3 + (b)3 + 3(2a)(b) (2a + b)
= (2a + b)3 | Using Identity VI
= (2a + b)(2a + b)(2a + b)

QUESTION: 8

p3+ q3 + r3 – 3pqr = ?​

Solution:
QUESTION: 9

Factorize: 125a3 – 27b3 – 225a2b + 135ab2.​

Solution:
QUESTION: 10

Evaluate (104)3

Solution:

(104)³
=(100 + 4)³
Use ( a +b)³ = a³ + b³ + 3ab( a + b)
then,
( 100 + 4)³ = ( 100)³ +(4)³ +3(100)(4)(100+4)
=(10²)³ + 64 + 1200(104)
=1000000 + 64 + 124800
=1124864.

QUESTION: 11

Factorize: (x – y)3 + (y – z)3 + (z – x)3

Solution:
QUESTION: 12

Factorise: x7y + xy7

Solution:
QUESTION: 13

What is the value 83 – 33 (without solving the cubes)?​

Solution:

(8-3)((8)² + 8×3 + (3)²) = 5(64 + 24 + 9) = 5×97 = 485

QUESTION: 14

p3+ q3 = ?​

Solution:
QUESTION: 15

What is the value of 53 – 13(without solving cube)?

Solution:

53 - 13 can be solved using the identity ;

(a3 - b3) = (a - b) (a2 + b2 + ab)

53 - 13 = ( 5 - 1 ) ( 52 + 12 + 5*1)

=(4) (25 + 1 + 5)

= (4) (31)

= 124

QUESTION: 16

Factorize: 27x3 – 125y3

Solution:
QUESTION: 17

a3 – b3 equals​

Solution:

It is the identity equation In the fraction equation QUESTION: 18

Evaluate: 303 + 203 – 503.​

Solution:

We have to find the value of QUESTION: 19

If , the value of x3 – y3 is :

Solution:
QUESTION: 20

Factorization of 103²-9 yields

Solution: