Ta có 

x + y = 2

=> (x+y)^2 = 4

=> x^2 + 2xy + y^2 = 4 

=> 10 + 2xy= 4

=> 2xy = -6

=> xy= -3

x^3 + y^3 = ( x+Y) ( x^2 - xy + y^2) = 2 ( 10 -- 3) = 2( 10  + 3 ) = 2.13 = 26

a. (a-b)^2 = (a-b)(a-b) = a^2 - ab - ba + b^2 = a^2 - 2ab + b^2

b. (a+b)^3= (a+b)(a+b)(a+b) = (a^2 + 2ab + b^2)(a + b) = a^3 + a^2b + 2a^2b + 2ab^2 + ab^2 + b^3 = a^3 + 3a^2b + 3b^2a + b^3

c. (a-b)^3= (a - b)(a-b)(a-b) = (a^2 - 2ab + b^2)(a - b) = a^3 - a^2b - 2a^2b + 2ab^2 + b^2a - b^3 = a^3 - 3a^2b + 3ab^2 - b^3

e. (a-b) ( a^2 + ab +b^2) = a^3 + a^2b + b^2a - ba^2 - ab^2 - b^3 = a^3 - b^3

g. ( a-b) ( a+b) = a^2 +ab -ab - b^2 = a^2 - b^2

\(a^3+b^3=c\left(3ab-c^2\right)\Rightarrow a^3+b^3+c^3-3abc=0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Rightarrow\left(a+b+c\right)\left[2a^2+2b^2+2c^2-2ab-2bc-2ca\right]=0\)

\(\Rightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)

\(\Rightarrow\orbr{\begin{cases}a+b+c=0\left(loai\right)\\a=b=c\end{cases}}\)

Mà a + b + c = 3 nên a = b = c = 1

Khi đó \(A=672.\left(1+1+1\right)+2=672.3+2=2018\)

Ta có A=\(\left(ab+bc+ca\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)

=\(2\left(a+b+c\right)+\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}-\frac{ab}{c}-\frac{bc}{a}-\frac{ca}{b}=2\left(a+b+c\right)\)

\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2=a^2-ab+b^2+3ab\left(1-2ab\right)+6a^2b^2\)

=\(\left(a+b\right)^2-3ab+3ab-6a^2b^2+6a^2b^2=1\)

2) Ta có \(A=\left(a-1\right)\left(b-1\right)\left(c-1\right)=abc-ab-bc-ca+a+b+c-1=0\)

bài 3 : Ta có \(A=\left(x-y\right)\left(x^2+xy+y^2\right)-36xy=12\left(x^2+xy+y^2\right)-36xy=12\left(x^2-2xy+y^2\right)\)

\(=12\left(x-y\right)^2=12.12^2=1728\)