Question

Use Euclid's division lemma to show that cube of any positive integer is either of the form 9m, 9m + 1 or 9m + 8 for some integer 'm'.

Answer 1

Let $a$ and $b$ be two positive integers, and $a>b$

$a=(b\times q)+r$ where $q$ and $r$ are positive integers and

$0\le r<b$

Let $b=3$ (If 9 is multiplied by 3 a perfect cube number is obtained)

$a=3q+r$ where $0\le r<3$

(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) $a=3q+1$

$a^3=(3q+1{)}^3$ $\left[\right(a+b{)}^3=a^3+b^3+3a^{2}b+3a{b}^2]$

$=27q^3+1+27q^2+9q=27q^3+27q^2+9q+1$

$=9(3q^3+3q^2+q)+1=9m+1$

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.

$\therefore$ cube of any positive integer is either of the form $9m,9m+1$ or $9m+8$ for some integer m.