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

To do :

We have to show that the cube of any positive integer is of the form $9m, 9m + 1$ or $9m + 8$.

Solution :

By Euclid's division algorithm,

$a = b q + r$, where $0 \leq r < b$.

Let a be the positive integer and b $=$ 3.

Now, $a = 3 q + r$, where $0 \leq r < 3$.

The possibilities of r is 0 , 1 , 2.

When $r = 0$,

$a = 3q$

Cubing both the sides,

$a^3 = (3q)^3$

$a^3 = 27q^3$

$a^3 = 9 (3q^3)$

$a^3 = 9m$ where $m = 3q^3$

When $r = 1$,

$a = 3q + 1$

Cubing both the sides,

$a^3 = (3q + 1)^3$

$a^3 = (3q)^3 + 1^3 + 3(3q)(1)(3q + 1)$

$a^3 = 27q^3 + 1 + 9q (3q + 1)$

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

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

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

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

When $r = 2$,

$a = 3q + 2$

Cubing both the sides,

$a^3 = (3q + 2)^3$

$a^3 = (3q)^3 + 2^3 + 3(3q)(2)(3q + 2)$

$a^3 = 27q^3 + 8 + 54q^2 + 36q$

$a^3 = 27q^3 + 54q^2 + 36q + 8$

$a^3 = 9 (3q^3 + 6q^2 + 4q) + 8$

$a^3 = 9m + 8$ where $m = (3q^3 + 6q^2 + 4q)$

Therefore, $a$ can be any of the form $9m$ or $9m + 1$ or, $9m + 8$.

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Updated on: 10-Oct-2022

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