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Test: Algebraic Identities Cubic Type - Class 9 MCQ


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

Test: Algebraic Identities Cubic Type for Class 9 2024 is part of Mathematics (Maths) Class 9 preparation. The Test: Algebraic Identities Cubic Type questions and answers have been prepared according to the Class 9 exam syllabus.The Test: Algebraic Identities Cubic Type MCQs are made for Class 9 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Algebraic Identities Cubic Type below.
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Test: Algebraic Identities Cubic Type - Question 1

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

Detailed Solution for Test: Algebraic Identities Cubic Type - Question 1

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

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

= 24h + 5h - 32 - 10

= 29h - 42, the simplification

Test: Algebraic Identities Cubic Type - Question 2

Evaluate (11)3

Detailed Solution for Test: Algebraic Identities Cubic Type - Question 2

(a+b)3 = a3 + b3 + 3ab(a + b)
11³= (10+1)³=1000+1+30(11)=1001+330=1331

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Test: Algebraic Identities Cubic Type - Question 3

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

Detailed Solution for Test: Algebraic Identities Cubic Type - Question 3

 

 

Test: Algebraic Identities Cubic Type - Question 4

Evaluate: 53 – 23 – 33

Detailed Solution for Test: Algebraic Identities Cubic Type - Question 4

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

Test: Algebraic Identities Cubic Type - Question 5

Factorise : 8a3+ b3 + 12a2b + 6a b2

Detailed Solution for Test: Algebraic Identities Cubic Type - Question 5

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)

Test: Algebraic Identities Cubic Type - Question 6

Evaluate (x+y+z)2

Detailed Solution for Test: Algebraic Identities Cubic Type - Question 6

just apply the formula of (a+b+c)2 = a2+b2+c2+2ab+2bc+2ca

Test: Algebraic Identities Cubic Type - Question 7

Factorise : 8a3 + b3 + 12a2 b + 6ab2

Detailed Solution for Test: Algebraic Identities Cubic Type - Question 7

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)

Test: Algebraic Identities Cubic Type - Question 8

p3+ q3 + r3 – 3pqr = ?​

Detailed Solution for Test: Algebraic Identities Cubic Type - Question 8

(p + q + r)(p2 + q2 + r2 – pq – qr - pr)

 

We know the cubic identity according to which,

p3+ q3 + r3 – 3pqr = (p + q + r)(p2 + q2 + r2 – p q – q r - pr)

Test: Algebraic Identities Cubic Type - Question 9

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

Detailed Solution for Test: Algebraic Identities Cubic Type - Question 9
- To factorize 125a^3 - 27b^3 - 225a^2b + 135ab^2, we recognize it as a cubic expression.
- First, notice the structure: it resembles a difference of cubes combined with other terms.
- Group terms: 125a^3 - 27b^3 can be factored as (5a - 3b)(25a^2 + 15ab + 9b^2).
- The remaining terms can be rearranged and factored using (3a - 5b).
- The complete factorization results in (3a - 5b)(5a - 3b)(3a + 5b)^2.
Test: Algebraic Identities Cubic Type - Question 10

Evaluate (104)3

Detailed Solution for Test: Algebraic Identities Cubic Type - Question 10

(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.

Test: Algebraic Identities Cubic Type - Question 11

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

Detailed Solution for Test: Algebraic Identities Cubic Type - Question 11

Test: Algebraic Identities Cubic Type - Question 12

Factorise: x7y + xy7

Detailed Solution for Test: Algebraic Identities Cubic Type - Question 12

Test: Algebraic Identities Cubic Type - Question 13

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

Detailed Solution for Test: Algebraic Identities Cubic Type - Question 13

Test: Algebraic Identities Cubic Type - Question 14

p3+ q3 = ?​

Test: Algebraic Identities Cubic Type - Question 15

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

Detailed Solution for Test: Algebraic Identities Cubic Type - Question 15

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

Test: Algebraic Identities Cubic Type - Question 16

Factorize: 27x3 – 125y3

Test: Algebraic Identities Cubic Type - Question 17

a3 – b3 equals​

Detailed Solution for Test: Algebraic Identities Cubic Type - Question 17

It is the identity equation

In the fraction equation

Test: Algebraic Identities Cubic Type - Question 18

Evaluate: 303 + 203 – 503.​

Detailed Solution for Test: Algebraic Identities Cubic Type - Question 18

We have to find the value of

Test: Algebraic Identities Cubic Type - Question 19

If  , the value of x3 – y3 is :

Detailed Solution for Test: Algebraic Identities Cubic Type - Question 19
  1. Create a common denominator:

    • Multiply both sides of the equation by xy to get rid of the fractions:
      • x² + y² = -xy
  2. Rearrange the equation:

    • Add xy to both sides:
      • x² + xy + y² = 0
  3. Substitute the values in:  x³ - y³ = (x - y)(x² + xy + y²)

  4.            We also know x² + xy + y² = 0 from step 2.

                Substitute these values into the identity:
    • x³ - y³ = (x-y)(0) = 0

Therefore, the value of x³ - y³ is 0.

So, the correct answer is option 3.

Test: Algebraic Identities Cubic Type - Question 20

Factorization of 103²-9 yields

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