Question

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

Answer 1

(i)
We have to find the cube of 35 using column method. We have: $a=3\text{and}b=5$

Column I a3	Column II $\mathbf{3}\mathbf{\times }{\mathit{a}}^{\mathbf{2}}\mathbf{\times }\mathit{b}$	Column III $\mathbf{3}\mathbf{\times }\mathit{a}\mathbf{\times }{\mathit{b}}^{\mathbf{2}}$	Column IV b3
$3^3=27$	$\mathbf{3}\mathbf{\times }{\mathit{a}}^{\mathbf{2}}\mathbf{\times }\mathit{b}\mathbf{=}3\times 3^2\times 5=135$	$\mathbf{3}\mathbf{\times }\mathit{a}\mathbf{\times }{\mathit{b}}^{\mathbf{2}}\mathbf{=}3\times 3\times 5^2=225$	$5^3=125$
+15	+23	+ 12	125
42	158	237
42	8	7	5

Thus, cube of 35 is 42875.

(ii)
We have to find the cube of 56 using column method. We have: $a=5\text{and}b=6$

Column I a3	Column II $\mathbf{3}\mathbf{\times }{\mathit{a}}^{\mathbf{2}}\mathbf{\times }\mathit{b}$	Column III $\mathbf{3}\mathbf{\times }\mathit{a}\mathbf{\times }{\mathit{b}}^{\mathbf{2}}$	Column IV b3
$5^3=125$	$\mathbf{3}\mathbf{\times }{\mathit{a}}^{\mathbf{2}}\mathbf{\times }\mathit{b}\mathbf{=}3\times 5^2\times 6=450$	$\mathbf{3}\mathbf{\times }\mathit{a}\mathbf{\times }{\mathit{b}}^{\mathbf{2}}\mathbf{=}3\times 5\times 6^2=540$	$6^3=216$
+50	+56	+ 21	216
175	506	561
175	6	1	6

Thus, cube of 56 is 175616.

(iii)
We have to find the cube of 72 using column method. We have: $a=7\text{and}b=2$

Column I a3	Column II $\mathbf{3}\mathbf{\times }{\mathit{a}}^{\mathbf{2}}\mathbf{\times }\mathit{b}$	Column III $\mathbf{3}\mathbf{\times }\mathit{a}\mathbf{\times }{\mathit{b}}^{\mathbf{2}}$	Column IV b3
$7^3=343$	$\mathbf{3}\mathbf{\times }{\mathit{a}}^{\mathbf{2}}\mathbf{\times }\mathit{b}\mathbf{=}3\times 7^2\times 2=294$	$\mathbf{3}\mathbf{\times }\mathit{a}\mathbf{\times }{\mathit{b}}^{\mathbf{2}}\mathbf{=}3\times 7\times 2^2=84$	$2^3=8$
+30	+8	+0	8
373	302	84
373	2	4	8

Thus, cube of 72 is 373248.