Verified answer

Ответ:

Применяем формулы:  [tex]\bf x^3+y^3=(x+y)(x^2-xy+y^2)[/tex]  и  

[tex]\bf x^2-y^2=(x-y)(x+y)[/tex]

[tex]1)\ \ \underbrace{(a+2)(a^2-2a+4)}_{a^3+2^3}-(a^2-2)\, a=a^3+8-(a^3-2a)=\boldsymbol {2a+8=2(a+4)}\\\\\\2)\ \ \underbrace{(a^3-2b^3)(a^3+2b^3)}_{(a^3)^2-(2b^3)^2}+\underbrace{(a^2+2b^2)(a^4-2a^2b^2+4b^4)}_{(a^2)^3+(2b^2)^3}=\\\\\\=a^6-4b^6+a^6+27b^6=\bf 2a^6+23b^6[/tex]

[tex]3)\ \ \underbrace{(3a^2+2)(9a^4-6a^2+4)}_{(3a^2)^3+2^3}-3\underbrace{(3a^3-1)(3a^3+1)}_{(3a^3)^2-1^2}=\\\\\\=27a^6+8-3(9a^6-1)=8+3=\bf 11[/tex]

[tex]4)\ \ \underbrace{(2+3a^2)(4-6a^2+9a^4)}_{2^3+(3a^2)^3}-3\underbrace{(3a^2+2)(3a^2-2)}_{(3a^2)^2-2^2}}=\\\\\\=8+27a^6-3\, (9a^4-4)=8+27a^6-27a^6+12=8+12=\bf 20[/tex]