show that cube of any positive integer is of the form 4m, 4m+1, 4m+3 for some integer m
Answers
Step-by-step explanation:
given positive integers are in the form of 4m , 4 m + 1 , 4 m + 3 .
by Euclid division Lemma we have
a=bq+r , 0 less than or equal to r less than b
here b= 4
r = 0,1,2,
if r=0,4 m +0 = 4m⇒(a)^3 = (4m)^3
=64m^3
=4(16m^3)
= 4m(m= 16m^3)
if r = 1,4m + 1 ⇒ (4m + 1)^3
= (4m)^3 + 3(4m)^2(1)+3(4m)(1)^2+1^3
= 64m^3 + 48m^2 + 12m + 1
= 4(16m^3 + 8m^2 + 3m) + 1
=4m + 1(16m^3 + 8m^2 + 3m)
if r = 2,4m + 2⇒(4m + 2)^3
= (4m)^3+3(4m)^2(2)+3(4m)(2)^2+2^3
= 64m^3+96m^2+48m+8
= 4(16m^3+24m^2+12m)+8
hence the cube of any positive integer is in the form of 4m , 4 m + 14m + 8
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