Find the cubes of the following numbers by column method :

(i) 35
(ii) 56
(iii) 72.

To do:

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

Solution:

(i)

I COLUMNII COLUMNIII COLUMNIV 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 COLUMNII 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 COLUMNII 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

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