Squares and Square Roots, Cubes and Cube Roots — Exercise 3(F)

How many perfect squares lie between 120 and 300?

1. 5

2. 6

3. 7

4. 8

Answer

We know that 11² = 121 and 17² = 289, which lie between 120 and 300.

Also, 10² = 100 < 120 and 18² = 324 > 300.

So, the perfect squares lying between 120 and 300 are :

121, 144, 169, 196, 225, 256 and 289

These are 7 in number.

Hence, option 3 is the correct option.

$\sqrt{41 - \sqrt{21 + \sqrt{19 - \sqrt{9}}}}$ = ?

1. 3

2. 5

3. 6

4. 6.4

Answer

Evaluating,

$\Rightarrow \sqrt{41 - \sqrt{21 + \sqrt{19 - \sqrt{9}}}}\\[1em] \Rightarrow \sqrt{41 - \sqrt{21 + \sqrt{19 - 3}}}\\[1em] \Rightarrow \sqrt{41 - \sqrt{21 + \sqrt{16}}}\\[1em] \Rightarrow \sqrt{41 - \sqrt{21 + 4}}\\[1em] \Rightarrow \sqrt{41 - \sqrt{25}}\\[1em] \Rightarrow \sqrt{41 - 5}\\[1em] \Rightarrow \sqrt{36}\\[1em] \Rightarrow 6$

Hence, option 3 is the correct option.

The value of $\sqrt{0.01} + \sqrt{0.81} + \sqrt{1.44} + \sqrt{0.0009}$ is

1. 2.03

2. 2.1

3. 2.11

4. 2.23

Answer

Evaluating,

⇒ $\sqrt{0.01} + \sqrt{0.81} + \sqrt{1.44} + \sqrt{0.0009}$

⇒ 0.1 + 0.9 + 1.2 + 0.03

⇒ 2.23

Hence, option 4 is the correct option.

If $\dfrac{52}{x} = \sqrt{\dfrac{169}{289}}$, then the value of $x$ is

1. 52

2. 58

3. 62

4. 68

Answer

Solving,

$\Rightarrow \dfrac{52}{x} = \sqrt{\dfrac{169}{289}}\\[1em] \Rightarrow \dfrac{52}{x} = \dfrac{\sqrt{169}}{\sqrt{289}}\\[1em] \Rightarrow \dfrac{52}{x} = \dfrac{13}{17}\\[1em] \Rightarrow 13x = 52 \times 17\\[1em] \Rightarrow x = \dfrac{52 \times 17}{13}\\[1em] \Rightarrow x = 4 \times 17\\[1em] \Rightarrow x = 68$

Hence, option 4 is the correct option.

? ÷ $\sqrt{0.25}$ = 25

1. 12.5

2. 25

3. 50

4. 125

Answer

Let the required number be $x$.

$\Rightarrow x \div \sqrt{0.25} = 25\\[1em] \Rightarrow x \div 0.5 = 25\\[1em] \Rightarrow x = 25 \times 0.5\\[1em] \Rightarrow x = 12.5$

Hence, option 1 is the correct option.

For what value of $\star$ the statement $\left(\dfrac{\star}{15}\right)\left(\dfrac{\star}{135}\right) = 1$ is true?

1. 15

2. 25

3. 35

4. 45

Answer

Let the value of $\star$ be $x$.

$\Rightarrow \left(\dfrac{x}{15}\right)\left(\dfrac{x}{135}\right) = 1\\[1em] \Rightarrow \dfrac{x^2}{2025} = 1\\[1em] \Rightarrow x^2 = 2025\\[1em] \Rightarrow x = \sqrt{2025}\\[1em] \Rightarrow x = \sqrt{(3 \times 3) \times (3 \times 3) \times (5 \times 5)}\\[1em] \Rightarrow x = 3 \times 3 \times 5\\[1em] \Rightarrow x = 45$

Hence, option 4 is the correct option.

What percentage of the numbers from 1 to 50 have squares that end in the digit 1?

1. 10

2. 5

3. 11

4. 20

Answer

We know that 1² = 1 and 9² = 81, so the square of a number ends in 1 only when the number ends in 1 or 9.

The numbers from 1 to 50 ending in 1 are 1, 11, 21, 31 and 41.

The numbers from 1 to 50 ending in 9 are 9, 19, 29, 39 and 49.

∴ Number of such numbers = 5 + 5 = 10

Required percentage = $\dfrac{10}{50}$ x 100%

Required percentage = 20%

Hence, option 4 is the correct option.

Which of the following cannot be the unit digit of a perfect square number?

1. 1

2. 6

3. 8

4. 9

Answer

According to the property of perfect squares, a number ending in 2, 3, 7 or 8 is never a perfect square.

So, 8 cannot be the unit digit of a perfect square number.

Hence, option 3 is the correct option.

$\dfrac{\sqrt{288}}{\sqrt{128}}$ = ?

1. $\dfrac{\sqrt{3}}{2}$

2. $\dfrac{3}{\sqrt{2}}$

3. $\dfrac{3}{2}$

4. 1.49

Answer

Solving,

$\Rightarrow \dfrac{\sqrt{288}}{\sqrt{128}}\\[1em] \Rightarrow \sqrt{\dfrac{288}{128}}\\[1em] \Rightarrow \sqrt{\dfrac{\cancel{32} \times 9}{\cancel{32} \times 4}}\\[1em] \Rightarrow \sqrt{\dfrac{9}{4}}\\[1em] \Rightarrow \dfrac{\sqrt{9}}{\sqrt{4}}\\[1em] \Rightarrow \dfrac{3}{2}$

Hence, option 3 is the correct option.

$\sqrt{72} \times \sqrt{98}$ = ?

1. 42

2. 64

3. 74

4. 84

Answer

Solving,

$\Rightarrow \sqrt{72} \times \sqrt{98}\\[1em] \Rightarrow \sqrt{72 \times 98}\\[1em] \Rightarrow \sqrt{2 \times 2 \times 2 \times 3 \times 3 \times 2 \times 7 \times 7}\\[1em] \Rightarrow \sqrt{(2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (7 \times 7)}\\[1em] \Rightarrow 2 \times 2 \times 3 \times 7\\[1em] \Rightarrow 84$

Hence, option 4 is the correct option.

The least number by which 294 must be multiplied to make it a perfect square, is

1. 2

2. 3

3. 6

4. 24

Answer

Resolving 294 into prime factors, we get :

$\begin{array}{l|r} 2 & 294 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$

294 = (7 × 7) × 2 × 3

Since the prime factors 2 and 3 are left unpaired, so 294 should be multiplied by 2 × 3, i.e., 6.

Hence, option 3 is the correct option.

The least number by which 1470 must be divided to get a number which is a perfect square, is

1. 5

2. 6

3. 15

4. 30

Answer

Resolving 1470 into prime factors, we get :

$\begin{array}{l|r} 2 & 1470 \\ \hline 3 & 735 \\ \hline 5 & 245 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$

1470 = (7 × 7) × 2 × 3 × 5

Since the prime factors 2, 3 and 5 are left unpaired, so 1470 should be divided by 2 × 3 × 5, i.e., 30.

Hence, option 4 is the correct option.

What is the least number to be added to 7700 to make it a perfect square?

1. 44

2. 77

3. 98

4. 131

Answer

Let us try to find the square root of 7700.

Clearly, (87)² < 7700 < (88)².

∴ Required number to be added = (88)² - 7700 = 7744 - 7700 = 44.

Hence, option 1 is the correct option.

$\sqrt{110.25} \times \sqrt{0.01} \div \sqrt{0.0025} - \sqrt{420.25}$ equals

1. 0.50

2. 0.64

3. 0.73

4. 0.75

Answer

Solving,

⇒ $\sqrt{110.25} \times \sqrt{0.01} \div \sqrt{0.0025} - \sqrt{420.25}$

⇒ 10.5 × 0.1 ÷ 0.05 - 20.5

⇒ 1.05 ÷ 0.05 - 20.5

⇒ 21 - 20.5

⇒ 0.5

Hence, option 1 is the correct option.

The greatest four digit perfect square number is

1. 9000

2. 9801

3. 9900

4. 9981

Answer

Greatest number of four digits = 9999.

Let us try to find the square root of 9999.

Clearly, (99)² is less than 9999 by 198.

So, the least number to be subtracted is 198.

Required number = 9999 - 198 = 9801

Hence, option 2 is the correct option.

A gardener plants 17956 trees in such a way that there are as many rows as there are trees in a row. The number of trees in a row is

1. 134

2. 136

3. 144

4. 154

Answer

Let the number of rows be $x$.

Then, the number of trees in each row = number of rows = $x$.

∴ Total number of trees = $x \times x = x^2$.

But, total number of trees = 17956 (given).

$\Rightarrow x^2 = 17956\\[1em] \Rightarrow x = \sqrt{17956}$

Marking periods and using the division method, we have :

⇒ $x = 134$

Hence, option 1 is the correct option.

$\sqrt[3]{4\dfrac{12}{125}}$ = ?

1. $1\dfrac{2}{5}$

2. $1\dfrac{3}{5}$

3. $1\dfrac{4}{5}$

4. $2\dfrac{2}{5}$

Answer

Solving,

$\Rightarrow \sqrt[3]{4\dfrac{12}{125}}\\[1em] \Rightarrow \sqrt[3]{\dfrac{512}{125}}\\[1em] \Rightarrow \dfrac{\sqrt[3]{512}}{\sqrt[3]{125}}\\[1em] \Rightarrow \dfrac{\sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2)}}{\sqrt[3]{5 \times 5 \times 5}}\\[1em] \Rightarrow \dfrac{2 \times 2 \times 2}{5}\\[1em] \Rightarrow \dfrac{8}{5}\\[1em] \Rightarrow 1\dfrac{3}{5}$

Hence, option 2 is the correct option.

By what least number must 21600 be multiplied so as to make it a perfect cube?

1. 6

2. 10

3. 20

4. 30

Answer

Resolving 21600 into prime factors, we get :

$\begin{array}{l|r} 2 & 21600 \\ \hline 2 & 10800 \\ \hline 2 & 5400 \\ \hline 2 & 2700 \\ \hline 2 & 1350 \\ \hline 3 & 675 \\ \hline 3 & 225 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$

21600 = (2 × 2 × 2) × (3 × 3 × 3) × 2 × 2 × 5 × 5

Since 2 × 2 and 5 × 5 are left after grouping the prime factors in triplets, so 21600 must be multiplied by 2 × 5, i.e., 10.

⇒ 21600 x 10 = 216000

⇒ 216000 = (2 x 2 x 2) x (2 x 2 x 2) x (3 x 3 x 3) x (5 x 5 x 5), which is a perfect cube.

Hence, option 2 is the correct option.

What is the smallest number by which 3600 be divided to make it a perfect cube?

1. 9

2. 50

3. 300

4. 450

Answer

Resolving 3600 into prime factors, we get :

$\begin{array}{l|r} 2 & 3600 \\ \hline 2 & 1800 \\ \hline 2 & 900 \\ \hline 2 & 450 \\ \hline 3 & 225 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$

3600 = (2 × 2 × 2) × 2 × 3 × 3 × 5 × 5

Since 2, 3 × 3 and 5 × 5 are left after grouping the prime factors in triplets, so 3600 must be divided by 2 × 3 × 3 × 5 × 5, i.e., 450.

⇒ 3600 ÷ 450 = 8

⇒ 8 = (2 × 2 × 2), which is a perfect cube.

Hence, option 4 is the correct option.

Fill in the blanks:

(i) There are ............ perfect squares between 1 and 100.

(ii) There are ............ perfect cubes between 1 and 100.

(iii) A perfect square number cannot have ......., ......., ....... and ....... as the unit digit.

(iv) The cube of a number ending in 7 will end in the digit ............

(v) 1 m² = ............ cm² and 1 m³ = ............ cm³.

(vi) (100)³ has ............ zeroes.

Answer

(i) The perfect squares lying between 1 and 100 are 4, 9, 16, 25, 36, 49, 64 and 81, which are 8 in number.

Hence, there are 8 perfect squares between 1 and 100.

(ii) The perfect cubes lying between 1 and 100 are 8, 27 and 64, which are 3 in number.

Hence, there are 3 perfect cubes between 1 and 100.

(iii) According to the property of perfect squares, a number ending in 2, 3, 7 or 8 is never a perfect square.

Hence, a perfect square number cannot have 2, 3, 7 and 8 as the unit digit.

(iv) We know that 7³ = 343, which ends in the digit 3.

For example, 17³ = 4913 and 27³ = 19683, which end in the digit 3.

Hence, the cube of a number ending in 7 will end in the digit 3.

(v) 1 m = 100 cm

$\Rightarrow 1 \text{ m}^2 = 100 \times 100 = 10000 \text{ cm}^2\\[1em] \Rightarrow 1 \text{ m}^3 = 100 \times 100 \times 100 = 1000000 \text{ cm}^3$

Hence, 1 m² = 10000 cm² and 1 m³ = 1000000 cm³.

(vi) (100)³ = 100 × 100 × 100 = 1000000

The number 1000000 has 6 zeroes.

Hence, (100)³ has 6 zeroes.

Write true (T) or false (F):

(i) The sum of two perfect squares is a perfect square.

(ii) The product of two perfect squares is a perfect square.

(iii) The square of every natural number is always greater than the number itself.

(iv) There exists no natural number whose square is equal to its cube.

(v) The cube root of 80000 is 200.

Answer

(i) False

Reason — The numbers 4 and 9 are perfect squares but 4 + 9 = 13, which is not a perfect square.

(ii) True

Reason — Let the two perfect squares be $a^2$ and $b^2$. Then, $a^2 \times b^2 = (ab)^2$, which is a perfect square. For example, 4 × 9 = 36 = 6².

(iii) False

Reason — For the natural number 1, 1² = 1, which is equal to the number itself and not greater than it.

(iv) False

Reason — For the natural number 1, 1² = 1 and 1³ = 1. So, the square of 1 is equal to its cube.

(v) False

Reason — (200)³ = 200 × 200 × 200 = 8000000 and not 80000. Also, 80000 = (2 × 2 × 2) × (2 × 2 × 2) × 2 × (5 × 5 × 5) × 5 is not a perfect cube, so its cube root is not a whole number.

Assertion (A): There are 42 natural numbers between 21² and 22².

Reason (R): There are $2n$ non-perfect square numbers between the squares of the numbers $n$ and $(n + 1)$.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

2. Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).

3. Assertion (A) is true but Reason (R) is false.

4. Assertion (A) is false but Reason (R) is true.

Answer

We know that there are $2n$ non-perfect square numbers between the squares of two consecutive numbers $n$ and $(n + 1)$.

So, Reason (R) is true.

Here, $n = 21$

Natural numbers between 21² and 22² = 2 × 21 = 42

So, Assertion (A) is true and Reason (R) is the correct explanation of Assertion (A).

Hence, option 1 is the correct option.

Assertion (A): For an integer $a, a^3$ is always greater than $a^2$.

Reason (R): Square of a negative integer is positive and cube of a negative integer is negative.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

2. Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).

3. Assertion (A) is true but Reason (R) is false.

4. Assertion (A) is false but Reason (R) is true.

Answer

Let $a = -2$.

$\Rightarrow a^2 = (-2)^2 = 4\\[1em] \Rightarrow a^3 = (-2)^3 = -8$

Here, $a^3$ is less than $a^2$.

So, Assertion (A) is false.

The square of a negative integer is positive as the product of two negative integers is positive and the cube of a negative integer is negative as the product of three negative integers is negative.

So, Reason (R) is true.

Hence, option 4 is the correct option.

Assertion (A): The value of $\sqrt[3]{27} + \sqrt[3]{0.008} + \sqrt[3]{0.064}$ is 3.6.

Reason (R): We cannot find the cube roots of negative numbers.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

2. Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).

3. Assertion (A) is true but Reason (R) is false.

4. Assertion (A) is false but Reason (R) is true.

Answer

Evaluating,

$\Rightarrow \sqrt[3]{27} + \sqrt[3]{0.008} + \sqrt[3]{0.064}\\[1em] \Rightarrow \sqrt[3]{3 \times 3 \times 3} + \sqrt[3]{0.2 \times 0.2 \times 0.2} + \sqrt[3]{0.4 \times 0.4 \times 0.4}\\[1em] \Rightarrow 3 + 0.2 + 0.4\\[1em] \Rightarrow 3.6$

So, Assertion (A) is true.

The cube of a negative number is negative, so every negative number has a cube root which is negative.

For example, $\sqrt[3]{-27} = -3$.

So, Reason (R) is false.

Hence, option 3 is the correct option.

If $\sqrt{21}$ = 4.578, then, the value of $\sqrt{\dfrac{3}{7}}$ is:

1. 0.654

2. 6.54

3. 1.526

4. none of these

Answer

Solving,

$\Rightarrow \sqrt{\dfrac{3}{7}}\\[1em] \Rightarrow \sqrt{\dfrac{3 \times 7}{7 \times 7}}\\[1em] \Rightarrow \sqrt{\dfrac{21}{49}}\\[1em] \Rightarrow \dfrac{\sqrt{21}}{\sqrt{49}}\\[1em] \Rightarrow \dfrac{4.578}{7}\\[1em] \Rightarrow 0.654$

Hence, option 1 is the correct option.

If $\sqrt{256} \div \sqrt{x} = 2$, then the value of $x$ is:

1. 8

2. 64

3. 32

4. 16

Answer

Solving,

$\Rightarrow \sqrt{256} \div \sqrt{x} = 2\\[1em] \Rightarrow \dfrac{16}{\sqrt{x}} = 2\\[1em] \Rightarrow \sqrt{x} = \dfrac{16}{2}\\[1em] \Rightarrow \sqrt{x} = 8\\[1em] \Rightarrow x = 8^2\\[1em] \Rightarrow x = 64$

Hence, option 2 is the correct option.

$\sqrt{248 + \sqrt{52 + \sqrt{144}}}$ is equal to:

1. 8

2. 12

3. 16

4. 20

Answer

Evaluating,

$\Rightarrow \sqrt{248 + \sqrt{52 + \sqrt{144}}}\\[1em] [\because \sqrt{144} = 12]\\[1em] \Rightarrow \sqrt{248 + \sqrt{52 + 12}}\\[1em] \Rightarrow \sqrt{248 + \sqrt{64}}\\[1em] [\because \sqrt{64} = 8]\\[1em] \Rightarrow \sqrt{248 + 8}\\[1em] \Rightarrow \sqrt{256}\\[1em] \Rightarrow 16$

Hence, option 3 is the correct option.

For integers $x$ and $y$, if $y = x^2$, then $x = \sqrt{y}$.

1. True

2. False

3. Cannot say anything

4. None of these

Answer

Let $x = -5$.

$\Rightarrow y = x^2 = (-5)^2 = 25\\[1em] \Rightarrow \sqrt{y} = \sqrt{25} = 5$

Here, $x = -5$ but $\sqrt{y}$ = 5.

So, $x$ is not equal to $\sqrt{y}$ when $x$ is a negative integer.

Hence, option 2 is the correct option.

There are 24 non square numbers between two consecutive perfect squares. The sum of these two consecutive perfect squares is:

1. 300

2. 305

3. 310

4. 313

Answer

We know that there are $2n$ non-perfect square numbers between the squares of two consecutive numbers $n$ and $(n + 1)$.

⇒ $2n = 24$

⇒ $n = 12$

So, the two consecutive perfect squares are 12² and 13².

⇒ 12² + 13²

⇒ 144 + 169

⇒ 313

Hence, option 4 is the correct option.

The cube of a given number can be smaller than the number itself. Is it true?

1. Yes

2. No

3. Can't say anything

4. None of these

Answer

Let the number be 0.5.

⇒ (0.5)³ = 0.5 × 0.5 × 0.5 = 0.125

Here, 0.125 < 0.5, so the cube is smaller than the number itself.

Again, let the number be -2.

⇒ (-2)³ = -8

Here, -8 < -2, so the cube is smaller than the number itself.

Hence, option 1 is the correct option.

Product of three distinct non-cube numbers can't be a perfect cube. This statement is:

1. True

2. False

3. Can't say anything

4. None of these

Answer

Let the three distinct numbers be 4, 6 and 9.

None of the numbers 4, 6 and 9 is a perfect cube.

⇒ 4 x 6 x 9 = 216

⇒ 216 = (2 x 2 x 2) x (3 x 3 x 3) = 6³, which is a perfect cube.

So, the product of three distinct non-cube numbers can be a perfect cube.

Hence, option 2 is the correct option.

If $x$ and $y$ are distinct positive rational numbers such that $x \lt \dfrac{1}{x}$ and $y \gt \dfrac{1}{y}$, then $x^3 - y^3$ will be:

1. positive

2. negative

3. neither positive nor negative

4. none of these

Answer

Given, $x$ is a positive rational number such that $x \lt \dfrac{1}{x}$.

$\Rightarrow x^2 \lt 1\\[1em] \Rightarrow x \lt 1$

So, $x^3 \lt 1$.

Given, $y$ is a positive rational number such that $y \gt \dfrac{1}{y}$.

$\Rightarrow y^2 \gt 1\\[1em] \Rightarrow y \gt 1$

So, $y^3$ > 1.

Since $x^3 \lt 1$ and $y^3 \gt 1$, so $x^3 \lt y^3$.

∴ $x^3 - y^3$ is negative.

Hence, option 2 is the correct option.

If $x^2 = 4 \times 27 \times 108 \times 256$, then $\sqrt[3]{x}$ is equal to:

1. 8

2. 1728

3. 12

4. 144

Answer

Solving,

Resolving each number into prime factors,

4 = 2², 27 = 3³, 108 = 2² × 3³ and 256 = 2⁸

$\Rightarrow x^2 = 4 \times 27 \times 108 \times 256\\[1em] \Rightarrow x^2 = 2^2 \times 3^3 \times (2^2 \times 3^3) \times 2^8\\[1em] \Rightarrow x^2 = 2^{(2 + 2 + 8)} \times 3^{(3 + 3)}\\[1em] \Rightarrow x^2 = 2^{12} \times 3^6\\[1em] \Rightarrow x = \sqrt{2^{12} \times 3^6}\\[1em] \Rightarrow x = 2^6 \times 3^3\\[1em] \Rightarrow x = 64 \times 27\\[1em] \Rightarrow x = 1728$

Now, finding the cube root of 1728,

$\Rightarrow \sqrt[3]{x}\\[1em] \Rightarrow \sqrt[3]{1728}\\[1em] \Rightarrow \sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3)}\\[1em] \Rightarrow 2 \times 2 \times 3\\[1em] \Rightarrow 12$

Hence, option 3 is the correct option.

$x$ and $y$ are two numbers ($x, y$ ≠ 1) such that $x$ is a perfect cube and $y$ is a perfect square. Which of the following numbers is a perfect cube?

(i) $x \times y$

(ii) $x \times y^2$

(iii) $x^2 \times y^2$

(iv) $x^2 \times y^3$

1. only (ii)

2. only (i) and (ii)

3. only (iv)

4. only (i), (ii) and (iii)

Answer

Since $x$ is a perfect cube, let $x = a^3$.

Since $y$ is a perfect square, let $y = b^2$.

Taking $x = 8$ and $y = 9$,

(i) $x \times y = 8 \times 9 = 72$, which is not a perfect cube.

(ii) $x \times y^2 = 8 \times 81 = 648$, which is not a perfect cube.

(iii) $x^2 \times y^2 = 64 \times 81 = 5184$, which is not a perfect cube.

(iv) $x^2 \times y^3 = (a^3)^2 \times (b^2)^3 = a^6 \times b^6 = (a^2b^2)^3$, which is a perfect cube.

For example, $x^2 \times y^3 = 64 \times 729 = 46656 = (36)^3$, which is a perfect cube.

Hence, option 3 is the correct option.