Задание 395

Решение:

• \({\largeа)}\ (a^3+1)(a^6-a^3+1)=(a^3+1)((a^3)^2-a^3\cdot1+1^2)=(a^3)^3+1^3=a^9+1\)

• \({\largeб)}\ (2+n^2)(n^4-2n^2+4)=(n^2+2)((n^2)^2-2n^2+2^2)=(n^2)^3+2^3=n^6+8\)

• \({\largeв)}\ (x+y^2)(x^2-xy^2+y^4)=(x+y^2)(x^2-xy^2+(y^2)^2)=x^3+(y^2)^3=x^3+y^6\)

• \({\largeг)}\ (p^3+q^2)(q^4-p^3q^2+p^6)=(p^3+q^2)((p^3)^2-p^3q^2+(q^2)^2)=(p^3)^3+(q^2)^3=p^9+q^6\)

• \({\largeд)}\ (a^4b^2-2a^2b+4)(2+a^2b)=(a^2b+2)((a^2b)^2-2a^2b+2^2)=(a^2b)^3+2^3=a^6b^3+8\)

• \({\largeе)}\ (9n^2-3nm+m^2)(m+3n)=(3n+m)((3n)^2-3nm+m^2)=(3n)^3+m^3=27n^3+m^3\)

• \({\largeж)}\ (3x+y)(9x^2-3xy+y^2)=(3x+y)((3x)^2-3xy+y^2)=(3x)^3+y^3=27x^3+y^3\)

• \({\largeз)}\ (a^4+1)(a^8-a^4+1)=(a^4+1)((a^4)^2-a^4\cdot1+1^2)=(a^4)^3+1^3=a^{12}+1\)

• \({\largeи)}\ (4x^4y^2-6x^2ya+9a^2)(3a+2x^2y)=(3a+2x^2y)((3a)^2-3a2x^2y+(2x^2y)^2)=(3a)^3+(2x^2y)^3=27a^3+8x^6y^3\)

• \({\largeк)}\ (5p^3+2q^2)(4q^4-10p^3q^2+25p^6)=(5p^3+2q^2)((5p^3)^2-5p^32q^2+(2q^2)^2)=(5p^3)^3+(2q^2)^3=125p^9+8q^6\)