Solution:
A number is a perfect cube only when each factor in the prime factorization is grouped in triples. Using this concept, the smallest number can be identified.
[i] 81
81 = 3 × 3 × 3 × 3
= 33 × 3
Here, the prime factor 3 is not grouped as a triplet. Hence, we divide 81 by 3, so that the obtained number becomes a perfect cube.
Thus, 81 ÷ 3 = 27 = 33 is a perfect cube.
Hence the smallest number by which 81 should be divided to make a perfect cube is 3.
[ii] 128
128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
= 23 × 23 × 2
Here, the prime factor 2 is not grouped as a triplet. Hence, we divide 128 by 2, so that the obtained number becomes a perfect cube.
Thus, 128 ÷ 2 = 64 = 43 is a perfect cube.
Hence the smallest number by which 128 should be divided to make a perfect cube is 2.
[iii] 135
135 = 3 × 3 × 3 × 5
= 33 × 5
Here, the prime factor 5 is not a triplet. Hence, we divide 135 by 5, so that the obtained number becomes a perfect cube.
135 ÷ 5 = 27 = 33 is a perfect cube.
Hence the smallest number by which 135 should be divided to make a perfect cube is 5.
[iv] 192
192 = 2 × 2 × 2 × 2 × 2 × 2 × 3
= 23 × 23 × 3
Here, the prime factor 3 is not grouped as a triplet. Hence, we divide 192 by 3, so that the obtained number becomes a perfect cube.
192 ÷ 3 = 64 = 43 is a perfect cube
Hence the smallest number by which 192 should be divided to make a perfect cube is 3.
[v] 704
704 = 2 × 2 × 2 × 2 × 2 × 2 × 11
= 23 × 23 × 11
Here, the prime factor 11 is not grouped as a triplet. Hence, we divide 704 by 11, so that the obtained number becomes a perfect cube.
Thus, 704 ÷ 11 = 64 = 43 is a perfect cube
Hence the smallest number by which 704 should be divided to make a perfect cube is 11.
☛ Check: NCERT Solutions for Class 8 Maths Chapter 7
Video Solution:
Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube [i] 81 [ii] 128 [iii] 135 [iv] 192 [v] 704
NCERT Solutions for Class 8 Maths Chapter 7 Exercise 7.1 Question 3
Summary:
The smallest number by which each of the following numbers must be divided to obtain a perfect cube [i] 81 [ii] 128 [iii] 135 [iv] 192 [v] 704 are [i] 3, [ii] 2, [iii] 5, [iv] 3, and [v] 11
☛ Related Questions:
- Which of the following numbers are not perfect cubes?[i] 216 [ii] 128 [iii] 1000 [iv] 100 [v] 46656
- Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.[i] 243 [ii] 256 [iii] 72 [iv] 675 [v] 100
- Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?
- Find the cube root of each of the following numbers by prime factorization method. [i] 64 [ii] 512 [iii] 10648 [iv] 27000 [v] 15625 [vi] 13824 [vii] 110592 [viii] 46656 [ix] 175616 [x] 91125
The smallest natural number by which 1296 be divided to get a perfect cube is _________
Answer
Verified
Hint: Here in this question we have to determine by which number the given number 1296 is divided such that it will be a perfect cube. Usually a prime factorization method is used to determine the cube root of a number, so the same method is implemented here also. On finding the prime factors of a number, then we can determine the smallest number which 1296 can be divided.
Complete step by step solution:
A perfect cube is a number which is equal to the number, multiplied
by itself, three times. If x is a perfect cube of y, then \[x = {y^3}\] .Therefore, if we take out the cube root of a perfect cube, we get a natural number and not a fraction. Hence, \[\sqrt[3]{x} = y\]
The cube root can be determined by using the prime factorisation. In the prime factorisation method we are going to divide the given number by the prime number. Then we are making a three same number as a pair, then it can be written in the form of a cube. As we know that the cube and cube
root are inverse to each other it will cancel. Hence we can determine the cube root of the given number.
Now we will consider the given question.
Here we have to determine the smallest number which should be divided by 1296, such that it will be a perfect cube.
Consider the number 1296
We use the prime factorisation method. So we have
\[
\,\,2\left| \!{\underline {\,
{1296} \,}} \right. \\
\,\,2\left| \!{\underline {\,
{648}
\,}} \right. \\
\,\,2\left| \!{\underline {\,
{324} \,}} \right. \\
\,\,2\left| \!{\underline {\,
{162} \,}} \right. \\
\,\,3\left| \!{\underline {\,
{81} \,}} \right. \\
\,\,3\left| \!{\underline {\,
{27} \,}} \right. \\
\,\,3\left| \!{\underline {\,
9 \,}} \right. \\
\,\,3\left| \!{\underline {\,
{3\,\,} \,}} \right.
\, \\
\,\,\,\,\,\,1 \;
\]
Therefore the number 1296 can be written as
\[ \Rightarrow 1296 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3\]
The number 2 is multiplied four times and the 3 is multiplied four times. So these can be written in the form of an exponent. so we have
\[ \Rightarrow 1296 = {2^4} \times {3^4}\]
This can be written as
\[ \Rightarrow 1296 = 2 \times {2^3} \times 3 \times {3^3}\]
We retain 2 cubes and 3 cubes,
and we multiply the number 2 and 3. So we have
\[ \Rightarrow 1296 = {2^3} \times {3^3} \times 6\]
By using the law of indices \[{a^m}.{b^m} = {[a.b]^m}\]
\[ \Rightarrow 1296 = {6^3} \times 6\]
When you divide the number 1296 by 6 then it will be a perfect cube.
So, the correct answer is “6”.
Note: We can verify the answer. On dividing the number 1296 by 6. The quotient will be 216. Using the prime factorization method we determine the factors.
\[
\,\,2\left|
\!{\underline {\,
{216} \,}} \right. \\
\,\,2\left| \!{\underline {\,
{108} \,}} \right. \\
\,\,2\left| \!{\underline {\,
{54} \,}} \right. \\
\,\,3\left| \!{\underline {\,
{27} \,}} \right. \\
\,\,3\left| \!{\underline {\,
9 \,}} \right. \\
\,\,3\left| \!{\underline {\,
{3\,\,} \,}} \right. \, \\
\,\,\,\,\,\,1 \;
\]
Therefore the number 216 can be written as
\[ \Rightarrow 216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3\]
The number 2 is multiplied thrice and the 3 is multiplied thrice, so we have
\[ \Rightarrow 216 = {2^3} \times {3^3} = {\left[ 6 \right]^3}\]
Therefore 216 is a perfect cube.