# Ex.7.2 Q3 Cubes and Cube Roots - NCERT Maths Class 8

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## Question

Q3. You are told that $$1,331$$ is a perfect cube. Can you guess without factorization what its cube root is?

Similarly, guess the cube roots of $$4913,\; 12167, \;32768.$$

## Text Solution

Reasoning:

By grouping the digits of cube into $$3$$ and using Table 7.2

Steps:

(i) $$1331$$

Step 1:

$$1 =$$ Group $$2$$

$$331 =$$ 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’s 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  $$\sqrt[\mathrm{3}]{1331}=11$$

(ii) $$4913$$

Step 1:

$$4 =$$ Group $$2$$

$$913 =$$ 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 it’s cube root ends in $$7$$. So, we get $$7$$ at the one’s place of the cube root. (Refer table 7.2 INFERENCE)

Step 3:From Group 2, which is 4.

\begin{align}{{1}^{3}}<4<{{2}^{3}}\end{align}

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

So, we get .\begin{align}\sqrt[\mathrm{3}]{4913}=17\end{align}

(iii) $$12167$$

Step 1:

$$12 =$$ Group $$2$$

$$167 =$$ Group $$1$$

Step 2:  From group $$1$$, which is $$167$$.

$$167$$ $$=$$ One’s digit is $$7$$

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

Step 3: From Group $$2$$, which is $$12$$

$${{2}^{3}}<12<{{3}^{3}}$$

Taking the lower limit. Therefore, ten’s digit of cube root is $$2$$.

So, we get .$$\sqrt[\mathrm{3}]{12167}=23$$

(iv) $$32768$$

Step 1:

$$32 =$$ Group $$2$$

$$768 =$$ Group $$1$$

Step 2: From group $$1$$, which is $$768$$.

$$768$$ $$=$$ One’s digit is $$8$$

We know that $$8$$ comes at the one’s place of a number only when it’s cube root ends in $$2$$. So, we get $$2$$ at the one’s place of the cube root. (Refer table 7.2 INFERENCE)

Step 3: From Group $$2$$, which is $$32$$.

$${{3}^{3}}<32<{{4}^{3}}$$

Taking lower limit. Therefore, ten’s digit of cube root is $$3$$.

So, we get .$$\sqrt[\mathrm{3}]{32768}=32$$

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