Use Euclid's division lemma to show that the cube of any positive integer is of the form 9 m, 9m+1 or 9m +8.
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Use Euclid's division lemma to show that the cube of any positive integer is of the form 9 m, 9m+1 or 9m +8.
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Let a and b be two positive integers, and a>b
a=(b×q)+r where q and r are positive integers and
0≤r<b
Let b=3 (If 9 is multiplied by 3 a perfect cube number is obtained)
a=3q+r where 0≤r<3
(i) if r=0,a=3q (ii) if r=1,a=3q+1 (iii) if r=2,a=3q+2
Consider, cubes of these
Case (i) a=3q
a
3
=(3q)
3
=27q
3
=9(3q
3
)=9m where m=3q
3
and 'm' is an integer.
Case (ii) a=3q+1
a
3
=(3q+1)
3
[(a+b)
3
=a
3
+b
3
+3a
2
b+3ab
2
]
=27q
3
+1+27q
2
+9q=27q
3
+27q
2
+9q+1
=9(3q
3
+3q
2
+q)+1=9m+1
where m=3q
3
+3q
2
+q and 'm' is an integer.
Case (iii) a=3q+2
a
3
=(3q+2)
3
=27q
3
+8+54q
2
+36q
=27q
3
−54q
2
+36q+8=9(3q
3
+6q
2
+4q)+8
9m+8, where m=3q
3
+6q
2
+4q and m is an integer.
∴ cube of any positive integer is either of the form 9m,9m+1 or 9m+8 for some integer m
Step-by-step explanation:
Note : with the help of cube formula we found the que ans
Cube formula :
[tex](a+b)^{3} = (a)^{3}+ (b )^{3} +3ab(a+b)\\ = (a)^{3} + (b )^{3} +3a^{2}b+3b^{2}a[/tex]
Let a be any +ve integer any number can be written in the form of
a = bq+r
b = 3
0 ≤ r ≥ 3
when r = 0
a = (3q+0)
Cube both side
[tex]a^{3} = (3q^{3})\\a^{3} = 27q^{3}\\a^{3}= 9[3q^{3}]\\a^{3} = 9m[/tex] where m = [tex]3q^{3}[/tex]
r= 1
a = (3q+1)
cube both side
[tex]a^{3}[/tex] = [tex](3q+1^{3})\\[/tex]
[tex]a^{3}[/tex] = [tex](3q^{3})[/tex]+ [tex]1^{3}[/tex] + 3×(3q)×(1) (3q+1)
= [tex](3q^{3})[/tex]+ [tex](1)^{3} + 3(3q)^{2} (1) + 3 (1)^{2} (3q)[/tex]
= [tex](27q^{3}) + 1 +9q^{2} + 9q[/tex]
= [tex]9[3q^{3}+q^{2}+q] + 1[/tex]
= 9m+1 where m = [tex]3q^{3}+q^{2}+q[/tex]
r = 2
a= 3q+2
Cube both side
[tex]a^{3}[/tex] = [tex](3q+2)^{3}[/tex]
[tex]a^{3} = (3q)^{3}+(2)^{3}+3(3q)^{2} (2) +3(2)^{2}(3q)\\ = 27q^{3} +8 +54q^{2} +36q\\ = 27q^{3} +54q^{2} +36q+8 \\ = 9 [3q^{3}+6q^{2} +49] +8 \\ = 9m+8[/tex]where m =[tex]3q^{3} +6q^{2}+4q[/tex]
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