Question

Prove that the value of the expression (125^2+25^2)(5^2–1) is divisible by 3

Answer 1

`(125^2+25^2)(5^2-1)=\\\\=(125^2+25^2)(25-1)=\\\\=(125^2+25^2)\cdot24=\\\\=(125^2+25^2)\cdot8\cdot{\bold{\underline3}}`

125 and 25 are integer ⇒125² and 25² are integer ⇒(125²+25²) is integer

One of factors of given expression is 3 so given expression is divisible by 3

q.e.d.

Answer 2

Explanation:

`A =(125^(2) + 25^(2) ) (5^(2) - 1)\\A = [(5^(3))^(2) + (5^(2))^(2) ] . (5^(2) - 1)\\A = (5^(6) + 5^(4) ). (5^(2) - 1)\\A = [5^(4) . ( 5^(2) + 5)].(5^(2) - 1) \\A = 5^(4) . (25+ 5). (5^(2) - 1)\\A = 5^(4) . 30. (5^(2) - 1)\\`

Since 30 is divisible by 3

Thus, A is divisible by 3

Good luck!