1, $a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)$

$=a^3+b^3+3a^3b+3ab^3+6a^2b^2$

$=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left(a^2+2ab+b^2\right)$

$=a^2-ab+b^2+3ab\left(a+b\right)^2$

$=a^2-ab+b^2+3ab$

$=a^2+2ab+b^2=\left(a+b\right)^2$

$=1$

Vậy A = 1

Bài 2: ( đặt đề bài là A )

Đặt $b+c-a=x,a+c-b=y,a+b-c=z$

$\Rightarrow a+b+c=x+y+z$

$\Leftrightarrow A=\left(x+y+z\right)^3-x^3-y^3-z^3$

$=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(x+z\right)-x^3-y^3-z^3$

$=3\left(x+y\right)\left(y+z\right)\left(x+z\right)$

$=3.2c.2a.2b=24abc$

Vậy...

Bài 3:

+) Xét p = 3 có: $p^2+2=11\in P$ ( t/m )

+) Xét $pe3$ thì:

+ $p=3k+1\Rightarrow p^2+2=\left(3k+1\right)^2+2=9k^2+6k+3⋮3otin P$

+ $p=3k+2\Rightarrow p^2+2=\left(3k+2\right)^2+2=9k^2+12k+6⋮3otin P$

Vậy p = 3

Bài 4:

$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2$

$\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4$

$\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=4$

$\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2c}{abc}+\dfrac{2a}{abc}+\dfrac{2b}{abc}=4$

$\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2\left(a+b+c\right)}{abc}=4$

$\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2=4$

$\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2$

$\Rightarrowđpcm$