Examples (Revised) - Chapter 7 - Cube & Cube Roots - Ncert Solutions class 8 - Maths

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Chapter 6 - Cube & Cube Roots | NCERT Solutions for Class 8 Maths

Example 1:

Is 243 a perfect cube?
Solution:

$243=\underline{3 \times 3 \times 3} \times 3 \times 3$

In the above factorisation $3 \times 3$ remains after grouping the 3 's in triplets. Therefore, 243 is not a perfect cube.

Example 2:

Is 392 a perfect cube? If not, find the smallest natural number by which 392 must be multiplied so that the product is a perfect cube.
Solution:

$392=\underline{2 \times 2 \times 2} \times 7 \times 7$
The prime factor 7 does not appear in a group of three. Therefore, 392 is not a perfect cube. To make its a cube, we need one more 7. In that case $392 \times 7=\underline{2 \times 2 \times 2} \times \underline{7 \times 7 \times 7}=2744 \quad$ which is a perfect cube.

Hence the smallest natural number by which 392 should be multiplied to make a perfect cube is 7 .

Example 3:

Is 53240 a perfect cube? If not, then by which smallest natural number should 53240 be divided so that the quotient is a perfect cube?
Solution:

$53240=\underline{2 \times 2 \times 2} \times \underline{11 \times 11 \times 11 \times 5}$
The prime factor 5 does not appear in a group of three. So, 53240 is not a perfect cube. In the factorisation 5 appears only one time. If we divide the number by 5 , then the prime factorisation of the quotient will not contain 5 .

So,
$
53240 \div 5=\underline{2 \times 2 \times 2 \times 11 \times 11 \times 11}
$

Hence the smallest number by which 53240 should be divided to make it a perfect cube is 5 .
The perfect cube in that case is $=10648$.
Example 4:

Is 1188 a perfect cube? If not, by which smallest natural number should 1188 be divided so that the quotient is a perfect cube?

Solution:

$1188=2 \times 2 \times \underline{3 \times 3 \times 3} \times 11$
The primes 2 and 11 do not appear in groups of three. So, 1188 is not a perfect cube. In the factorisation of 1188 the prime 2 appears only two times and the prime 11 appears once. So, if we divide 1188 by $2 \times 2 \times 11=44$, then the prime factorisation of the quotient will not contain 2 and 11

Hence the smallest natural number by which 1188 should be divided to make it a perfect cube is 44 .
And the resulting perfect cube is $1188 \div 44=27\left(=3^3\right)$.

Example 5:

Is 68600 a perfect cube? If not, find the smallest number by which 68600 must be multiplied to get a perfect cube.
Solution:

We have, $68600=2 \times 2 \times 2 \times 5 \times 5 \times 7 \times 7 \times 7$. In this factorisation, we find that there is no triplet of 5 .
So, 68600 is not a perfect cube. To make it a perfect cube we multiply it by 5 .

Thus,
$
\begin{aligned}
68600 \times 5 & =2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 7 \times 7 \times 7 \\
& =343000, \text { which is a perfect cube. }
\end{aligned}
$

Observe that 343 is a perfect cube. From Example 5 we know that 343000 is also perfect cube.

Example 6:

Find the cube root of 8000 .
Solution:

Prime factorisation of 8000 is $\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \times \underline{5 \times 5 \times 5}$
So, $\quad \sqrt[3]{8000}=2 \times 2 \times 5=20$
Example 7:

Find the cube root of 13824 by prime factorisation method.
Solution:

$
13824=\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \times \underline{3 \times 3 \times 3}=2^3 \times 2^3 \times 2^3 \times 3^3 .
$

Therefore, $\sqrt[3]{13824}=2 \times 2 \times 2 \times 3=24$