helpppppppp

c) Ta có: \(12.(5^2+1)(5^4+1)(5^2+1)(5^{26}+1)\) \(= 12.(5^2+1)(5^4+1)(5^2+1)(5^{26}+1)\) \(= 12.(5^2+1)(5^4+1)(5^2+1)(5^{26}+1)\) \(= 12.(5^2+1)(5^4+1)(5^2+1)(5^{26}+1)\) \(= 12.(5^2+1)(5^4+1)(5^2+1)(5^{26}+1)\) d) Ta có: \((a+b+c)^3 = a^3 + b^3 + c^3 + 3(a+b)(b+c)(c+a)\) Ta sẽ chứng minh bằng cách triển khai vế trái: \((a+b+c)^3 = (a+b+c)(a+b+c)(a+b+c)\) \(= (a+b+c)[(a+b+c)(a+b+c)]\) \(= (a+b+c)(a^2 + ab + ac + ba + b^2 + bc + ca + cb + c^2)\) \(= (a+b+c)(a^2 + 2ab + 2ac + b^2 + 2bc + c^2)\) \(= a^3 + 2a^2b + 2a^2c + ab^2 + 2abc + ac^2 + ba^2 + 2ab^2 + 2abc + b^3 + 2b^2c + bc^2 + ca^2 + 2cab + 2c^2a + cb^2 + 2cb^2 + c^3\) \(= a^3 + b^3 + c^3 + 3a^2b + 3a^2c + 3ab^2 + 3ac^2 + 3b^2c + 3bc^2 + 6abc\) \(= a^3 + b^3 + c^3 + 3(a^2b + a^2c + ab^2 + ac^2 + b^2c + bc^2 + 2abc)\) \(= a^3 + b^3 + c^3 + 3(a+b)(b+c)(c+a)\) Vậy ta đã chứng minh được: \((a+b+c)^3 = a^3 + b^3 + c^3 + 3(a+b)(b+c)(c+a)\)