use euclid division lemma to ehow that cube of any positive integer is of the form 7m ,7m+1 or 7m+6
Answers
Answer:
Step-by-step explanation:
Let 'a' be any positive integer and b = 7.
Using Euclid Division Lemma,
a = bq + r [ 0 ≤ r < b ]
⇒ a = 7q + r [ 0 ≤ r < 7 ]
Now, possible value of r :
r = 0, r = 1, r = 2, r = 3, r = 4, r = 5, r = 6
CASE I :
If we take, r = 0
⇒ a = 7q + 0
⇒ a = 7q
On cubing both sides,
⇒ a³ = (7q)³
⇒ a³ = 7 ( 49 q³ )
⇒ a³ = 7m [49q³ = m as integer]
CASE II :
If we take, r = 1
⇒ a = 7q + 1
On cubing both sides ;
⇒ a³ = (7q + 1)³
⇒ a³ = 343q³ + 1³ + 3 * 7q * 1 ( 7q + 1 )
⇒ a³ = 343q³ + 1 + 147q² + 21q
⇒ a³ = 7 ( 49q³ + 21q² + 3q ) + 1
⇒ a³ = 9m + 1 [ Take m as some integer ]
CASE III :
If we take r = 2,
⇒ a = 7q + 2
On cubing both sides ;
⇒ a³ = (7q + 2)³
⇒ a³ = 343q³ + 2³ + 3 * 7q * 2 ( 7q + 2 )
⇒ a³ = 343q³ + 8 + 294q² + 84q
⇒ a³ = 7 ( 49q³ + 42q² + 12q ) + 8
⇒ a³ = 7m + 8 [Take m as some integer]
And so on.
Hence, the cube of any positive integer is in the form of 7m, 7m+1 or 7m+8.
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Identity used ;
( a + b )³ = a³ + b³ + 3ab ( a + b )
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+
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
Step-by-step explanation: