# Find the cubes of the following numbers by column method :(i) 35(ii) 56(iii) 72.

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To do:

We have to find the cube of the given numbers by the column method.

Solution:

(i)

 I COLUMN II COLUMN III COLUMN IV COLUMN $a^3$ $3\times(a^2)\times(b)$ $3\times(a)\times(b^2)$ $b^3$ $3^3$ $3\times(3^2)\times(5)$ $3\times3\times(5^2)$ $(5^3)$ $=27$ $=135$ $=225$ $=125$ $+15$ $+23$ $+12$ $42$ $158$ $237$ $42$ $8$ $7$ $5$

Therefore,

$(35)^3=42875$

(ii)

 I COLUMN II COLUMN II COLUMN IV COLUMN $a^3$ $3\times(a^2)\times(b)$ $3\times(a)\times(b^2)$ $b^3$ $5^3$ $3\times(5^2)\times(6)$ $3\times5\times(6^2)$ $(6^3)$ $=125$ $=450$ $=540$ $216$ $+50$ $+56$ $+21$ $175$ $506$ $561$ $175$ $6$ $1$ $6$

Therefore,

$(56)^3=175616$

(iii)

 I COLUMN II COLUMN II COLUMN IV COLUMN $a^3$ $3\times(a^2)\times(b)$ $3\times(a)\times(b^2)$ $b^3$ $7^3$ $3\times(7^2)\times(2)$ $3\times7\times(2^2)$ $(2^3)$ $=343$ $=294$ $=84$ $8$ $+30$ $+8$ $373$ $302$ $373$ $2$ $4$ $8$

Therefore,

$(72)^3=373248$

Updated on 10-Oct-2022 12:46:26