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

## 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 .\(\sqrt[\mathrm{3}]{32768}=32\)