Chapter 3: Squares and Square Roots & Cubes and Cube Roots

Solution:-

False.

Solution:-

False.

So, factors of 363 × 81 are 3 × 11 × 11 × 3 × 3 × 3 × 3

= 3³ × 11 × 11 × 3 × 3

Solution:-

False.

Because, cube root of 8 is 2 only.

i.e. 2 × 2 × 2 = 8

Solution:-

False.

Let us consider LHS = ³√ (8 + 27)

= ³√ (35)

Then, RHS = ³√8 + ³√27

= 2 + 3

= 5

By comparing LHS and RHS

LHS ≠ RHS

³√ (35) ≠ 5

Solution:-

False.

Let us take -27

Cube root of -27 i.e. ³√-27 = ³√((-3) × (-3) × (-3))

= -3

Solution:-

False.

Let us take -3

Now, square of the -3 i.e. -3² = 9

Then, cube of the same number i.e. -3³ = -27

Solve the following questions.

Solution:-

The first five square numbers are,

1 × 1 = 1

2 × 2 = 4

3 × 3 = 9

4 × 4 = 16

5 × 5 = 25

Solution:-

The first three multiple of 3 are 3, 6 and 9

Then,

Cubes of first three multiples of 3,

= 3³ = 3 × 3 × 3 = 27

= 6³ = 6 × 6 × 6 = 216

= 9³ = 9 × 9 × 9 = 729

Solution:-

Factors of 500 are 2 × 2 × 5 × 5 × 5

Now,

Group the factors into pairs,

500 = (2 × 2) × (5 × 5 × 5)

= 2² × 5³

So, the prime factors 2 and 5 do not occur in pairs.

Therefore, 500 is not a perfect square.

Solution:-

The first nine consecutive odd numbers are 1, 3, 5, 7, 9, 11, 13, 15 and 17

Then,

We know that 81 is a perfect square,

81 = (9)²

Now,

The sum of first nine consecutive odd numbers,

= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17

= 81