Question
You are told that 1331 is a perfect cube. Can you guess without factorisation what its cube root is? Similarly, guess the cube roots of 4913, 12167, and 32768.
✓
Answer
(i) Yes, the cube root of 1331 can be guessed without factorization.
$10^3=1000 \text { and } 20^3=8000$
So, $10<\sqrt[3]{1331}<20$
Since 1331 ends in 1, its cube root ends in 1.
The number between 10 and 20 that ends in 1 is 11 .
$\therefore \sqrt[3]{1331}=11 .$
(ii) $\sqrt[3]{4913}$
$10^3=1000 \text { and } 20^3=8000$
So, $10<\sqrt[3]{4913}<20$
Since 4913 ends in 3 , its cube root ends in 7.
The number between 10 and 20 that ends in 7 is 17.
$\therefore \sqrt[3]{4913}=17 .$
(iii) $\sqrt[3]{12167}$
$20^3=8000 \text { and } 30^3=27000$
$\text { So, } 20<\sqrt[3]{12167}<30$
Since 12167 ends in 7 , its cube root ends in 3.
The number between 20 and 30 that ends in 3 is 23 .
$\therefore \sqrt[3]{12167}=23 .$
(iv) $\sqrt[3]{32768} $
$30^3=27000 \text { and } 40^3=64000$
$\text { So, } 30<\sqrt[3]{32768}<40$
Since 32768 ends in 8 , its cube root ends in 2.
The number between 30 and 40 that ends in 2 is 32 .
$\sqrt[3]{32768} =32 .$
$10^3=1000 \text { and } 20^3=8000$
So, $10<\sqrt[3]{1331}<20$
Since 1331 ends in 1, its cube root ends in 1.
The number between 10 and 20 that ends in 1 is 11 .
$\therefore \sqrt[3]{1331}=11 .$
(ii) $\sqrt[3]{4913}$
$10^3=1000 \text { and } 20^3=8000$
So, $10<\sqrt[3]{4913}<20$
Since 4913 ends in 3 , its cube root ends in 7.
The number between 10 and 20 that ends in 7 is 17.
$\therefore \sqrt[3]{4913}=17 .$
(iii) $\sqrt[3]{12167}$
$20^3=8000 \text { and } 30^3=27000$
$\text { So, } 20<\sqrt[3]{12167}<30$
Since 12167 ends in 7 , its cube root ends in 3.
The number between 20 and 30 that ends in 3 is 23 .
$\therefore \sqrt[3]{12167}=23 .$
(iv) $\sqrt[3]{32768} $
$30^3=27000 \text { and } 40^3=64000$
$\text { So, } 30<\sqrt[3]{32768}<40$
Since 32768 ends in 8 , its cube root ends in 2.
The number between 30 and 40 that ends in 2 is 32 .
$\sqrt[3]{32768} =32 .$
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