Question

Using the method of successive subtraction examine whether or not the following numbers are perfect cubes:
(i) 130
(ii) 345
(iii) 792
(iv) 1331

Answer 1

(i)
We have:

$$\begin{gathered} 130 \\ \underline{1} \\ 129 \\ \underline{7} \\ 122 \\ \underline{19} \\ 103 \\ \underline{37} \\ 66 \\ \underline{61} \\ 5 \end{gathered}$$

$\because$ The next number to be subtracted is 91, which is greater than 5.

$\therefore$ 130 is not a perfect cube.

(ii)
We have:

$$\begin{gathered} 345 \\ \underline{1} \\ 344 \\ \underline{7} \\ 337 \\ \underline{19} \\ 318 \\ \underline{37} \\ 281 \\ \underline{61} \\ 220 \\ \underline{91} \\ 129 \\ \underline{127} \\ 2 \end{gathered}$$

$\because$ The next number to be subtracted is 161, which is greater than 2.

$\therefore$ 345 is not a perfect cube.

(iii)
We have:

$$\begin{gathered} 792 \\ \underline{1} \\ 791 \\ \underline{7} \\ 784 \\ \underline{19} \\ 765 \\ \underline{37} \\ 728 \\ \underline{61} \\ 667 \\ \underline{91} \\ 576 \\ \underline{127} \\ 449 \\ \underline{169} \\ 280 \\ \underline{217} \\ 63 \end{gathered}$$

$\because$ The next number to be subtracted is 271, which is greater than 63.

$\therefore$ 792 is not a perfect cube.

(iv)
We have:

$$\begin{gathered} 1331 \\ \underline{1} \\ 1330 \\ \underline{7} \\ 1323 \\ \underline{19} \\ 1304 \\ \underline{37} \\ 1267 \\ \underline{61} \\ 1206 \\ \underline{91} \\ 1115 \\ \underline{127} \\ 988 \\ \underline{169} \\ 819 \\ \underline{217} \\ 602 \\ \underline{271} \\ 331 \\ \underline{331} \\ 0 \end{gathered}$$

$\because$ The subtraction is performed 11 times.

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

Thus, 1331 is a perfect cube.