Question

Use EDL to show that cube of any positive integer of the from 7mor7mt or 7m+6.

Answer 1

Step-by-step explanation:

# here is your answer //

Let take b = 7. (/here b<a. and a , b , q and r are the positive integers)

according to Euclid division lemma

a = bq+r. (0≥r>b)

a = 7q+r (0≥r>7)

so r values are

r -------->0,1,2,3,4,5,6

1) when r value equals to zero (r=0)

a = 7q+r

a = 7q+0

a = 7q

cube both side

a³ = 343q³

a³ = 7(49q³)

a³ = 7m. (/here m is positive integer and m = 49q³)

2) if value of r is equals to 1 (r=1)

a = 7q+r

a = 7q+1

cube both side

(a)³ = (7q+1)³

a³ = 343q³+1³+3(7q)(1)(7q+1)

a³. = 343q³+21q(7q+1) + 1

a³. = 343q³+ 147q² + 21q + 1

a³ = 7(49q³+21q²+3q)+1

a³ = 7m + 1. (/here m is positive integer and m = 49q³+21q²+3q)

3) if r value equals to 6 ( r = 6)

a = 7q+r

a = 7q+6

cube both side

(a)³ = (7q+6)³

a³ = 343q³+216+3(7q)(6)(7q+6)

a³ = 343q³+126q(7q+6)+216

a³ = 343q³+882q²+756q+216

a³ = 7(49q³+127q²+108q)+216

a³ = 7m+216. (/here m is positive integer and m = 49q³+127q²+108q)

I think 7m + 6 is not a cube of any positive integer

// hope it is helpful #R //☺☺

Answer 2

Step-by-step explanation:

Step-by-step explanation: 1) when r value equals to zero (r=0) a = 7q+r. a = 7q+0. a = 7q. cube both side. a³ = 343q³ a³ = 7(49q³) a³ = 7m. 2) if value of r is equals to 1 (r=1) a = 7q+r. a = 7q+1. cube both side. (a)³ = (7q+1)³ a³ = 343q³+1³+3(7q)(1)(7q+1) a³. = 343q³+21q(7q+1) + 1. 3) if r value equals to 6 ( r = 6)