Question

find the cube pf a no. without actual multiplication using the method of repeated addition no. are 5 (odd) 10(even)

Answer 1

The question is not completely clear. So, I am assuming that we need to find the cube of 5 and 10 through the method of repeated addition.

We know that for a number 'a', 
$a^{3}$ = a * a * a

a * a =  'a' added 'a' times

Using these properties, we solve the problem.

(1) $5^{3}$
= 5 * 5 * 5
= (5 * 5) * 5
= (5 + 5 + 5 + 5 + 5) * 5
= (5 + 5 + 5 + 5 + 5) + (5 + 5 + 5 + 5 + 5) + (5 + 5 + 5 + 5 + 5) + (5 + 5 + 5 + 5 + 5) + (5 + 5 + 5 + 5 + 5)
= 25 + 25 + 25 + 25 + 25
= 125

Therefore, $5^{3}$  = 125

(2) $10^{3}$ 
= 10 * 10 * 10
= (10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10) * 10
= (10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10) + (10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10) + (10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10) + (10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10) + (10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10) + (10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10) + (10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10) + (10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10) + (10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10) + (10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10)
= 100 + 100 + 100 + 100 + 100 + 100 + 100 + 100 + 100 + 100 
= 1000

Thus, $10^{3}$  = 1000

(Do let me know if I have got the question wrong, I will edit or delete the answer.)