Задание 437

Решение:

• \({\largeа)}\ (a+1)(a+2)(a^2+4)(a^2+1)(a-2)(a-1)=(a+1)(a-1)(a^2+1)(a+2)(a-2)(a^2+4)=(a^2-1)(a^2+1)(a^2-4)(a^2+4)=((a^2)^2-1^2)((a^2)^2-4^2)=(a^4-1)(a^4-16)=a^8\ -\) \(16a^4\) \(-\) \(a^4\) \(+\ 16=a^8-17a^4+16\)

• \({\largeб)}\ (a+b+c)(a+b-c)-2ab=(a+b)^2-c^2-2ab=a^2\ +\) \(2ab\) \(+\ b^2-c^2\ -\) \(2ab\) \(=a^2+b^2-c^2\)

• \({\largeв)}\ (a-b)(a+b)(b^2+a^2)(a^4+b^4)=(a^2-b^2)(a^2+b^2)(a^4+b^4)=((a^2)^2-(b^2)^2)(a^4+b^4)=(a^4-b^4)(a^4+b^4)=(a^4)^2-(b^4)^2=a^8-b^8\)

• \({\largeг)}\ (a+b)^3-3ab(a+b)=a^3\ +\) \(3a^2b\) \(+\) \(3ab^2\) \(+\ b^3\ -\) \(3a^2b\) \(-\) \(3ab^2\) \(=a^3+b^3\)

• \({\largeд)}\ 3ab(a-b)+(a-b)^3=\) \(3a^2b\) \(-\) \(3ab^2\) \(+\ a^3\ -\) \(3a^2b\) \(+\) \(3ab^2\) \(-\ b^3=a^3-b^3\)

• \({\largeе)}\ (a^2-2)(a^2+2)-(2-a^2)^2=(a^2)^2-2^2-(2^2-2\cdot2\cdot{a^2}+(a^2)^2)=\) \(a^4\) \(-\) \(4\) \(-\) \(4\) \(+\ 4a^2\ -\) \(a^4\) \(=4a^2-8\)