Question

if a and b are natural number s and a-b is divisible by 3 then a3-b3 is divisible by which number

Answer 1

a^3 -b^3 = (a-b)(a^2 + ab+ b^2)   -------> eq(i)

given that a-b is divisible by 3. so, (a-b)= 3(x)
 
If we put (a-b) = 3(x) in eq(i)
we get , 
a^3-b^3 =3(x)(a^2+ab+b^2)
 Therefore, a^3-b^3 is divisible by 3.

Answer 2

Given:

Two natural numbers a and b, a-b is divisible by 3.

To find:

$a^{3}-b^{3}$ is divisible by which number.

Solution:

Since, $a-b$ is divisible by 3, let, $a-b=3k$ for some integer k.

Square both sides of $a-b=3k$.

$(a-b)^2=(3k)^2$

Use the identity and expand the expression on the left-hand side.

$a^2-2ab+b^2=9k^2$

The identity $a^3-b^3$ is given by, $a^3-b^3=(a-b)(a^2+ab+b^2)$.

Re-arrange the terms inside the second bracket on the right-hand side of the above expression.

$a^3-b^3=(a-b)(a^2-2ab+b^2+3ab)$

Substitute $9k^2$ for $(a^2-2ab+b^2)$ and 3k for a-b into the above equation.

$a^3-b^3=3k(9k^2+3ab)\\=9k(3k^2+ab)$

Thus, it can be seen that $(3k^2+ab)$ is an integer, so, $9k(3k^2+ab)$ is divisible by 9.