Question

If n is a natural number, then by which number 9^2n

– 3^2n is always divisible.

Answer 1

${9}^{2n} - {3}^{2n} \\ = {( {9}^{n} )}^{2} - {( {3}^{n} )}^{2} \\ = ( {9}^{n} - {3}^{n})( {9}^{n} + {3}^{n})$
surely for any n, whether even or odd, the factor
$( {9}^{n} - {3}^{n}) \:$
can always be broken in the form:
$(9 - 3)(p)$
where p depends upon n, 9 and 3
so the statement
${9}^{2n} - {3}^{2n} \:$
is always divisible by 9 – 3, i.e., 6.

Hope you got help from my answer, if yes, then mark my answer as brainliest ☺️☺️.