Using Euclid division lemma, show that the cube of any positive integer is of the form 9 t or 9 t + 1 or 9 t +8, for some integer
Answers
Answer:
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:
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