Question

Show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8
Why this is in 3m+1 form plz explain

Answer 1

$\huge\bf\boxed{\boxed{\boxed{\red{\mathfrak{QUESTION \: : }}}}}$

$Show \: that \: the \: cube \: of \: any \\ \\ positive \: integer \: is \: of \: the \\ \\ form \: 9m, \: 9m + 1 \: or \: 9m + 8 \: .. \\ \\ Which \: of \: this \: is \: in \: 3k+1 \: form \: ???$

$\huge\bf\boxed{\boxed{\boxed{\orange{\mathfrak{ANSWER \: : }}}}}$

Let a be any positive integer

and b = 3

$\large\bf\boxed{a \: = \: 3q \: + \: r}$

where q ≥ 0 and 0 ≤ r < 3

Therefore, every number can be represented as these three forms.

There are three cases.

$\large\bf\boxed{CASE \: 1 \: :}$

When a = 3q,

Where m is an integer such that m = 1

$\large\bf\boxed{CASE \: 2 \: :}$

When a = 3q + 1,

a³ = (3q +1)³

a³ = 27q³+ 27q²+ 9q + 1

a³ = 9(3q³+ 3q² + q) + 1

a³ = 9m + 1

Where m is an integer such that

m = (3q³+ 3q²+ q)

$\large\bf\boxed{CASE \: 3 \: :}$

When a = 3q + 2,

a³ = (3q +2)³

a³ = 27q³ + 54q² + 36q + 8

a³ = 9(3q³ + 6q² + 4q) + 8

a³ = 9m + 8

Where m is an integer

such that

m = (3q³ + 6q² + 4q)

$\huge\bf\boxed{\boxed{\boxed{\green{\mathfrak{HENCE}}}}}$

the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

$\huge\bf\boxed{\boxed{\boxed{\blue{\mathfrak{ALSO}}}}}$

From the above,

9m + 1

is in the form of 3k + 1

where,

k = 3

Answer 2

Answer:

Let a be any positive integer

and b = 3

a=3q+r\large\bf\boxed{a \: = \: 3q \: + \: r}

a=3q+r

where q ≥ 0 and 0 ≤ r < 3

Therefore, every number can be represented as these three forms.

There are three cases.

CASE1:\large\bf\boxed{CASE \: 1 \: :}

CASE1:

When a = 3q,

Where m is an integer such that m = 1

CASE2:\large\bf\boxed{CASE \: 2 \: :}

CASE2:

When a = 3q + 1,

a³ = (3q +1)³

a³ = (3q +1)³

a³ = 27q³+ 27q²+ 9q + 1

a³ = 9(3q³+ 3q² + q) + 1

a³ = 9m + 1

Where m is an integer such that

m = (3q³+ 3q²+ q)

CASE3:\large\bf\boxed{CASE \: 3 \: :}

CASE3:

When a = 3q + 2,

a³ = (3q +2)³

a³ = 27q³ + 54q² + 36q + 8

a³ = 9(3q³ + 6q² + 4q) + 8

a³ = 9m + 8

Where m is an integer

such that

m = (3q³ + 6q² + 4q)

HENCE\huge\bf\boxed{\boxed{\boxed{\green{\mathfrak{HENCE}}}}}

HENCE

the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

ALSO\huge\bf\boxed{\boxed{\boxed{\blue{\mathfrak{ALSO}}}}}

ALSO

From the above,

9m + 1

is in the form of 3k + 1

where,

k =3 ans .........