**Question:**

Find the cube root of each of the following natural numbers:

(i) 343

(ii) 2744

(iii) 4913

(iv) 1728

(v) 35937

(vi) 17576

(vii) 134217728

(viii) 48228544

(ix) 74088000

(x) 157464

(xi) 1157625

(xii) 33698267

**Solution:**

(i)

Cube root using units digit:

Let us consider 343.

The unit digit is 3; therefore, the unit digit in the cube root of 343 is 7.

There is no number left after striking out the units, tens and hundreds digits of the given number; therefore, the cube root of 343 is 7.

Hence, $\sqrt[3]{343}=7$

(ii)

Cube root using units digit:

Let us consider 2744.

The unit digit is 4; therefore, the unit digit in the cube root of 2744 is 4.

After striking out the units, tens and hundreds digits of the given number, we are left with 2.

Now, 1 is the largest number whose cube is less than or equal to 2.

Therefore, the tens digit of the cube root of 2744 is 1.

Hence, $\sqrt[3]{2744}=14$

(iii)

Cube root using units digit:

Let us consider 4913.

The unit digit is 3; therefore, the unit digit in the cube root of 4913 is 7.

After striking out the units, tens and hundreds digits of the given number, we are left with 4.

Now, 1 is the largest number whose cube is less than or equal to 4.

Therefore, the tens digit of the cube root of 4913 is 1.

Hence, $\sqrt[3]{4913}=17$

(iv)

Cube root using units digit:

Let us consider 1728.

The unit digit is 8; therefore, the unit digit in the cube root of 1728 is 2.

After striking out the units, tens and hundreds digits of the given number, we are left with 1.

Now, 1 is the largest number whose cube is less than or equal to 1.

Therefore, the tens digit of the cube root of 1728 is 1.

Hence, $\sqrt[3]{1728}=12$

(v)

Cube root using units digit:

Let us consider 35937.

The unit digit is 7; therefore, the unit digit in the cube root of 35937 is 3.

After striking out the units, tens and hundreds digits of the given number, we are left with 35.

Now, 3 is the largest number whose cube is less than or equal to $35\left(3^{3}<35<4^{3}\right)$.

Therefore, the tens digit of the cube root of 35937 is 3.

Hence, $\sqrt[3]{35937}=33$

(vi)

Cube root using units digit:

Let us consider the number 17576.

The unit digit is 6; therefore, the unit digit in the cube root of 17576 is 6.

After striking out the units, tens and hundreds digits of the given number, we are left with 17.

Now, 2 is the largest number whose cube is less than or equal to $17\left(2^{3}<17<3^{3}\right)$.

Therefore, the tens digit of the cube root of 17576 is 2.

Hence, $\sqrt[3]{17576}=26$

(vii)

Cube root by factors:

On factorising 134217728 into prime factors, we get:

$134217728=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$ $\times 2 \times 2 \times 2 \times 2 \times 2$

On grouping the factors in triples of equal factors, we get:

$134217728=\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\}$$\times\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\}$

Now, taking one factor from each triple, we get:

$\sqrt[3]{134217728}=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2=512$

(viii)

Cube root by factors:

On factorising 48228544 into prime factors, we get:

$48228544=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 7 \times 7 \times 7 \times 13 \times 13 \times 13$

On grouping the factors in triples of equal factors, we get:

$48228544=\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times\{7 \times 7 \times 7\} \times\{13 \times 13 \times 13\}$

Now, taking one factor from each triple, we get:

$\sqrt[3]{48228544}=2 \times 2 \times 7 \times 13=364$

(ix)

Cube root by factors:

On factorising 74088000 into prime factors, we get:

$74088000=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5 \times 7 \times 7 \times 7$

On grouping the factors in triples of equal factors, we get:

$74088000=\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times\{3 \times 3 \times 3\} \times\{5 \times 5 \times 5\} \times\{7 \times 7 \times 7\}$

Now, taking one factor from each triple, we get:

$\sqrt[3]{74088000}=2 \times 2 \times 3 \times 5 \times 7=420$

(x)

Cube root using units digit:

Let is consider 157464.

The unit digit is 4; therefore, the unit digit in the cube root of 157464 is 4.

After striking out the units, tens and hundreds digits of the given number, we are left with 157.

Now, 5 is the largest number whose cube is less than or equal to $157\left(5^{3}<157<6^{3}\right)$

Therefore, the tens digit of the cube root 157464 is 5.

Hence, $\sqrt[3]{157464}=54$

(xi)

Cube root by factors:

On factorising 1157625 into prime factors, we get:

$1157625=3 \times 3 \times 3 \times 5 \times 5 \times 5 \times 7 \times 7 \times 7$

On grouping the factors in triples of equal factors, we get:

Now, taking one factor from each triple, we get:

$\sqrt[3]{1157625}=3 \times 5 \times 7=105$

(xii)

Cube root by factors:

On factorising 33698267 into prime factors, we get

$33698267=17 \times 17 \times 17 \times 19 \times 19 \times 19$

On grouping the factors in triples of equal factors, we get:

$33698267=\{17 \times 17 \times 17\} \times\{19 \times 19 \times 19\}$

Now, taking one factor from each triple, we get:

$\sqrt[3]{33698267}=17 \times 19=323$

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