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

Answer 1

(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 ( $3^3<35<4^3$).
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 ($2^3<17<3^3$).
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:
$134217728=\left\{2\times 2\times 2\right\}\times \left\{2\times 2\times 2\right\}\times \left\{2\times 2\times 2\right\}\times \left\{2\times 2\times 2\right\}\times \left\{2\times 2\times 2\right\}\times \left\{2\times 2\times 2\right\}\times \left\{2\times 2\times 2\right\}\times \left\{2\times 2\times 2\right\}\times \left\{2\times 2\times 2\right\}$

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:
$48228544=\left\{2\times 2\times 2\right\}\times \left\{2\times 2\times 2\right\}\times \left\{7\times 7\times 7\right\}\times \left\{13\times 13\times 13\right\}$

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:
$74088000=\left\{2\times 2\times 2\right\}\times \left\{2\times 2\times 2\right\}\times \left\{3\times 3\times 3\right\}\times \left\{5\times 5\times 5\right\}\times \left\{7\times 7\times 7\right\}$

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 ($5^3<157<6^3$).
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:
$1157625=\left\{3\times 3\times 3\right\}\times \left\{5\times 5\times 5\right\}\times \left\{7\times 7\times 7\right\}$

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:
$33698267=\left\{17\times 17\times 17\right\}\times \left\{19\times 19\times 19\right\}$

Now, taking one factor from each triple, we get:
$\sqrt[3]{33698267}=17\times 19=323$