Question

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

Answer 1

Step-by-step 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!

Answer 2

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

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.