Can you guess without factorisation what is the cube root of 32768

 By grouping the digits, we get 1 and 331

We know that, since, the unit digit of cube is 1, the unit digit of cube root is 1.

∴ We get 1 as unit digit of the cube root of 1331.

The cube of 1 matches with the number of second group.

∴ The ten's digit of our cube root is taken as the unit place of smallest number.

We know that, the unit’s digit of the cube of a number having digit as unit’s place 1 is 1.

$\therefore \sqrt[3]{1331}=11$

 By grouping the digits, we get 4 and 913

We know that, since, the unit digit of cube is 3, the unit digit of cube root is 7.

∴ we get 7 as unit digit of the cube root of 4913.

We know $1^{3}=1 \text { and } 2^{3}=8$ , 1 > 4 > 8.

Thus, 1 is taken as ten digit of cube root.

$\therefore \sqrt[3]{4913}=17$

 By grouping the digits, we get 12 and 167.

We know that, since, the unit digit of cube is 7, the unit digit of cube root is 3.

∴ 3 is the unit digit of the cube root of 12167

We know $2^{3}=8 \text { and } 3^{3}=27$ , 8 > 12 > 27.

Thus, 2 is taken as ten digit of cube root.

$\therefore \sqrt[3]{12167}=23$

 By grouping the digits, we get 32 and 768.

We know that, since, the unit digit of cube is 8, the unit digit of cube root is 2.

∴ 2 is the unit digit of the cube root of 32768.

We know $3^{3}=27 \text { and } 4^{3}=64$ , 27 > 32 > 64.

Thus, 3 is taken as ten digit of cube root.

$\therefore \sqrt[3]{32768}=32$

Solution:

By grouping the digits of the number into triplets starting from one's digit

(i) 1331

Step 1: 1 = Group 2 and 331 = Group 1

Step 2: From group 1, one’s digit of the cube root can be identified.

331= One’s digit is 1

Hence cube root one’s digit is 1.

Step 3: From group 2, which is 1 only.

Hence cube root’s ten’s digit is 1.

So, we get ∛1331 = 11.

(ii) 4913

Step 1: 4 = Group 2 and 913 = Group 1

Step 2: From group 1, which is 913.

913 = One’s digit is 3

We know that 3 comes at the one’s place of a number only when its cube root ends in 7. So, we get 7 at the one’s place of the cube root. (Refer to table 7.2 INFERENCE)

Step 3: From Group 2, which is 4.

13 < 4 < 23

Taking lower limit. Therefore, the ten’s digit of cube root is 1.

So, we get ∛1331 = 17

(iii) Similarly, we get ∛12167 = 23

(iv) Similarly, we get ∛32768 = 32

☛ Check: NCERT Solutions for Class 8 Maths Chapter 7

Video Solution:

NCERT Solutions for Class 8 Maths Chapter 7 Exercise 7.2 Question 3

Summary:

You are told that 1,331 is a perfect cube. The cube root of 1,331 is 11. Similarly, the cube roots of 4913, 12167, 32768 are 17, 23, and 32

☛ Related Questions:

Answer

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Hint: To solve the question, we need to know the concept of cubic root of a number. The cube root of a number is the number which when multiplied by itself three times gives the number itself. In this question, we will factorize the number to find the cubic root of the number. We will use the Prime Factorization to get the factors of the number as it will help in finding the factors of the number easily.

Complete step by step answer:

The question asks us to find the cubic root of 32768. To solve our problem, the first step would be to prime factorize our number which is equal to 32768.To find the cubic root of our number using the method of prime factorization, we will write the number as a product of its prime factors. The prime factorization of 32768 can be done as follows: $\begin{align} & 2\left| \!{\underline {\, 32768 \,}} \right. \\ & 2\left| \!{\underline {\, 16384 \,}} \right. \\ & 2\left| \!{\underline {\, 8192 \,}} \right. \\ & 2\left| \!{\underline {\, 4096 \,}} \right. \\ & 2\left| \!{\underline {\, 2048 \,}} \right. \\ & 2\left| \!{\underline {\, 1024 \,}} \right. \\ & 2\left| \!{\underline {\, 512 \,}} \right. \\ & 2\left| \!{\underline {\, 256 \,}} \right. \\ & 2\left| \!{\underline {\, 128 \,}} \right. \\ & 2\left| \!{\underline {\, 64 \,}} \right. \\ & 2\left| \!{\underline {\, 32 \,}} \right. \\ & 2\left| \!{\underline {\, 16 \,}} \right. \\ & 2\left| \!{\underline {\, 8 \,}} \right. \\ & 2\left| \!{\underline {\, 4 \,}} \right. \\ & 2\left| \!{\underline {\, 2 \,}} \right. \\ & \left| \!{\underline {\, 1 \,}} \right. \\ \end{align}$ Now that we have prime factorized 32768. On substituting 32768 with the product of its prime factors, we get the following equation:$\Rightarrow 32768=\left( 2\times 2\times 2 \right)\times \left( 2\times 2\times 2 \right)\times \left( 2\times 2\times 2 \right)\times \left( 2\times 2\times 2 \right)\times \left( 2\times 2\times 2 \right)$ Now taking cube roots on both side of the equation, we get:$\Rightarrow \sqrt[3]{32768}=\sqrt{\left( 2\times 2\times 2 \right)\times \left( 2\times 2\times 2 \right)\times \left( 2\times 2\times 2 \right)\times \left( 2\times 2\times 2 \right)\times \left( 2\times 2\times 2 \right)}$ Taking out the numbers which repeat three times out of the cubic root, we get our new equation as:$\begin{align} & \Rightarrow \sqrt[3]{32768}=2\times 2\times 2\times 2\times 2 \\ & \therefore \sqrt[3]{32768}=32 \\ \end{align}$ Thus, we get the final resulting number as 32.Hence, the cubic root of 32768 by the method of prime factorization comes out to be 32 .

Note: We can always check our solution after calculating the cubic root or any other root of a number. In this case, this can be done by finding the cube of the final number obtained and checking if its equal to our original number. In our case, the cube of 32 comes out to be equal to 32768. This verifies that our procedure and the final answer, both are correct.