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Cube of the following numbers are:

i) 8³= 8 x 8 x 8 = 512

ii) 16³= 16 x 16 x 16 = 4096

iii) 21³= 21 x 21 x 21= 9261

iv) 30³⁼ 30 x 30 x 30= 27000

Correct option is  B) 1331

Correct option is (B) 1331

\(11^3=11^2\times11=121\times11=1331.\)

125 = 5 × 5 × 5 = (5)³

Yes, 125 is a perfect cube. 

[∵ It can be written as product of 3 same numbers]

Correct option is  A) 3³

Correct option is (A) 3³

7+9+11 = 27 = \(3^3.\)

Perfect cube numbers between 1 and 100 = 8, 27, 64 

Perfect cube numbers between 1 and 500 = 8, 27, 64, 125, 216, 343 

Perfect cube numbers between 1 and 1000 = 8, 27, 64, 125, 216, 343, 512, 729

Perfect cubes between 500 and 1000 = 512 and 729

Correct option is  B) 3

Correct option is (B) 3

\(\because\) Unit's digit in 1997 is 7.

\(\therefore\) Unit's digit in \((1997)^3\) is unit's digit in \(7^3=3.\)

\((\because7^3=343\Rightarrow\) unit's digit in \(7^3\) is 3)

(i) 3375 

Step 1: Start making groups of three digits starting from the unit place. i.e.;

Step 2: First group is 375. Its units digit is 5. 

∴ The cube root is also ends with 5. 

∴ The units place of the cube root will be 5.

Step 3: Now take the second group,

 i.e., 3 we know that 1³ < 3³ <2³ 

∴ The least number is 1. 

∴ The tens digit of a cube root will be 1. 

∴ The required number = 15

 ∴ \(\sqrt[3]{3375}\) = \(\sqrt[3]{15\,\times\,15\,\times15}\) = \(\sqrt[3]{15}{^3}\) = 15

(ii) 5832 

Step 1: Start making groups of three digits starting from the unit place.

Step 2: The units digit of 832 is 2. 

∴ The cube root of the number ends with units digit 8. 

[∵ 8 x 8 x 8 = 512]

Step 3: In the second group i.e., 5 lie between 1 and 6

 i.e., 1³ < 5 < 2³ 

∴ The tens digit of a number will bel. 

∴ The required number is 18.

∴ \(\sqrt[3]{5832}\) = \(\sqrt[3]{18\,\times\,18\,\times18}\) = \(\sqrt[3]{18}{^3}\) = 18

Correct option is  A) 3375

Correct option is (A) 3375

\(15^3=15^2\times15=225\times15=3375.\)

Correct option is  C) 343

Correct option is (C) 343

\(\sqrt[3]p=7\)

\(\Rightarrow p=7^3=343.\)

Correct option is A) 1728

Correct option is (A) 1728

\(\sqrt[3]{x}\) = 12 \(\Rightarrow\) \((x)^{\frac13}=12\)

\(\Rightarrow\) \(x=12^3=12^2\times12\)

\(\Rightarrow\) \(x=144\times12\)

\(\Rightarrow\) x = 1728.

Correct option is  D) 4

Correct option is (D) 4

\(\sqrt[3]{64}=\sqrt[3]{4\times4\times4}=\sqrt[3]{4^3}\) \(=(4^3)^\frac13=(4)^{3\times\frac13}=4^1=4.\)

Correct option is  B) 3 abc

Correct option is (B) 3 abc

Since, \(a^3+b^3+c^3\) - 3ab \(=(a+b+c)(a^2+b^2+c^2-bc-ca-ab)\)

\(\therefore\) If a + b + c = 0, then \(a^3+b^3+c^3\) = 3 abc.

Correct option is  C) 3² 

Correct option is (C) 3²

\(1^3+2^3\) = 1+8 = 9 = \(3^2\).

Correct option is  C) 1729

Correct option is (C) 1729

\(12^3+1^3\) = 1728+1 = 1729 \((\because12^3=1728)\)

Correct option is  B) 6² 

Correct option is (B) 6²

\(1^3+2^3+3^3\) = 1+8+27 = 36 = \(6^2.\)

Correct option is   A) x = \(\sqrt[3]{y}\)

Correct option is B) 216

The two digited square and cubic number is 64 

∴ 64 = 8 x 8 = 8² 

⇒ \(\sqrt{64}\) = 8 

∴ 64 = 4 x 4 x 4 = 4³ 

⇒ \(\sqrt[3]{64}\) = 4

i) 512

Step 1: Start making groups of three digits starting from the unit place. 

i.e., \(\overline{512}\) First group is 512 

Step 2: First group i.e. 512 will give us the units digit of the cube root. As 512 ends with 2, then its cube root ends with 8 (2 x 2 x 2) So the units place of the cube root will be 8.

Step 3: Now take the second group i.e. 0.Which is 0³ < 1 < 2³ . 

So the least number is ‘0′. 

∴ Tens digit of a cube root of a number be 0. 

∴ \(\sqrt[3]{512}\) = 08 = 8

(ii) 2197 

Step 1: Start making groups of three digits starting from the unit place.

Step 2: First group i.e., 197 will give us the units digit of the cube root. 

As 197 ends with 7, its cube root ends with 3. ‘

 [∵ 3 x 3 x 3 = 27] 

∴ Its units digit is 7.

Step 3: Now take the second group i.e.,2 

We know that i3 < 2 < 2 

∴ The least number be 1. 

∴ The required number is 13.

∴ \(\sqrt[3]{2197}\) = \(\sqrt[3]{13\,\times\,13\,\times13}\) = \(\sqrt[3]{13}{^3}\) = 13

81 = 3 × 3 × 3 × 3 = 3⁴ 

No, 81 is not a perfect cube. 

[∵ 81 can’t be written as product of 3 same numbers.]

Correct option is  D) 64

Correct option is (D) 64

\(64=8^2\) and \(64=4^3\)

Thus, 64 is a perfect square and perfect cube number.