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Question

# Find the smallest number that must be subtracted from those of the numbers in question 2 which are not perfect cubes, to make them perfect cubes. What are the corresponding cube roots?

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Solution

## (i) We have: $130\phantom{\rule{0ex}{0ex}}\overline{)1}\phantom{\rule{0ex}{0ex}}129\phantom{\rule{0ex}{0ex}}\overline{)7}\phantom{\rule{0ex}{0ex}}122\phantom{\rule{0ex}{0ex}}\overline{)19}\phantom{\rule{0ex}{0ex}}103\phantom{\rule{0ex}{0ex}}\overline{)37}\phantom{\rule{0ex}{0ex}}66\phantom{\rule{0ex}{0ex}}\overline{)61}\phantom{\rule{0ex}{0ex}}5\phantom{\rule{0ex}{0ex}}$ $\because$ The next number to be subtracted is 91, which is greater than 5. $\therefore$ 130 is not a perfect cube. However, if we subtract 5 from 130, we will get 0 on performing successive subtraction and the number will become a perfect cube. If we subtract 5 from 130, we get 125. Now, find the cube root using successive subtraction. We have: $125\phantom{\rule{0ex}{0ex}}\overline{)1}\phantom{\rule{0ex}{0ex}}124\phantom{\rule{0ex}{0ex}}\overline{)7}\phantom{\rule{0ex}{0ex}}117\phantom{\rule{0ex}{0ex}}\overline{)19}\phantom{\rule{0ex}{0ex}}98\phantom{\rule{0ex}{0ex}}\overline{)37}\phantom{\rule{0ex}{0ex}}61\phantom{\rule{0ex}{0ex}}\overline{)61}\phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}$ $\because$ The subtraction is performed 5 times. $\therefore$ $\sqrt[3]{125}=5$ Thus, it is a perfect cube. (ii) We have: $345\phantom{\rule{0ex}{0ex}}\overline{)1}\phantom{\rule{0ex}{0ex}}344\phantom{\rule{0ex}{0ex}}\overline{)7}\phantom{\rule{0ex}{0ex}}337\phantom{\rule{0ex}{0ex}}\overline{)19}\phantom{\rule{0ex}{0ex}}318\phantom{\rule{0ex}{0ex}}\overline{)37}\phantom{\rule{0ex}{0ex}}281\phantom{\rule{0ex}{0ex}}\overline{)61}\phantom{\rule{0ex}{0ex}}220\phantom{\rule{0ex}{0ex}}\overline{)91}\phantom{\rule{0ex}{0ex}}129\phantom{\rule{0ex}{0ex}}\overline{)127}\phantom{\rule{0ex}{0ex}}2\phantom{\rule{0ex}{0ex}}$ $\because$ The next number to be subtracted is 161, which is greater than 2. $\therefore$ 345 is not a perfect cube. However, if we subtract 2 from 345, we will get 0 on performing successive subtraction and the number will become a perfect cube. If we subtract 2 from 345, we get 343. Now, find the cube root using successive subtraction. $343\phantom{\rule{0ex}{0ex}}\overline{)1}\phantom{\rule{0ex}{0ex}}342\phantom{\rule{0ex}{0ex}}\overline{)7}\phantom{\rule{0ex}{0ex}}335\phantom{\rule{0ex}{0ex}}\overline{)19}\phantom{\rule{0ex}{0ex}}316\phantom{\rule{0ex}{0ex}}\overline{)37}\phantom{\rule{0ex}{0ex}}279\phantom{\rule{0ex}{0ex}}\overline{)61}\phantom{\rule{0ex}{0ex}}218\phantom{\rule{0ex}{0ex}}\overline{)91}\phantom{\rule{0ex}{0ex}}127\phantom{\rule{0ex}{0ex}}\overline{)127}\phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}$ $\because$ The subtraction is performed 7 times. $\therefore$ $\sqrt[3]{343}=7$ Thus, it is a perfect cube. (iii) We have: $792\phantom{\rule{0ex}{0ex}}\overline{)1}\phantom{\rule{0ex}{0ex}}791\phantom{\rule{0ex}{0ex}}\overline{)7}\phantom{\rule{0ex}{0ex}}784\phantom{\rule{0ex}{0ex}}\overline{)19}\phantom{\rule{0ex}{0ex}}765\phantom{\rule{0ex}{0ex}}\overline{)37}\phantom{\rule{0ex}{0ex}}728\phantom{\rule{0ex}{0ex}}\overline{)61}\phantom{\rule{0ex}{0ex}}667\phantom{\rule{0ex}{0ex}}\overline{)91}\phantom{\rule{0ex}{0ex}}576\phantom{\rule{0ex}{0ex}}\overline{)127}\phantom{\rule{0ex}{0ex}}449\phantom{\rule{0ex}{0ex}}\overline{)169}\phantom{\rule{0ex}{0ex}}280\phantom{\rule{0ex}{0ex}}\overline{)217}\phantom{\rule{0ex}{0ex}}63\phantom{\rule{0ex}{0ex}}$ $\because$ The next number to be subtracted is 271, which is greater than 63. $\therefore$ 792 is not a perfect cube. However, if we subtract 63 from 792, we will get 0 on performing successive subtraction and the number will become a perfect cube. If we subtract 63 from 792, we get 729. Now, find the cube root using the successive subtraction. We have: $729\phantom{\rule{0ex}{0ex}}\overline{)1}\phantom{\rule{0ex}{0ex}}728\phantom{\rule{0ex}{0ex}}\overline{)7}\phantom{\rule{0ex}{0ex}}721\phantom{\rule{0ex}{0ex}}\overline{)19}\phantom{\rule{0ex}{0ex}}702\phantom{\rule{0ex}{0ex}}\overline{)37}\phantom{\rule{0ex}{0ex}}665\phantom{\rule{0ex}{0ex}}\overline{)61}\phantom{\rule{0ex}{0ex}}604\phantom{\rule{0ex}{0ex}}\overline{)91}\phantom{\rule{0ex}{0ex}}513\phantom{\rule{0ex}{0ex}}\overline{)127}\phantom{\rule{0ex}{0ex}}386\phantom{\rule{0ex}{0ex}}\overline{)169}\phantom{\rule{0ex}{0ex}}217\phantom{\rule{0ex}{0ex}}\overline{)217}\phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}$ $\because$ The subtraction is performed 9 times. $\therefore$ $\sqrt[3]{729}=9$ Thus, it is perfect cube.

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