Nếu \(a+b=2\) thì :

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

\(=2a^2+4ab+2b^2=2\left(a+b\right)^2=2.2^2=8\) (TMĐB)

Vậy \(a^3+b^3+6ab=8\) thì \(a+b=2\)

Ta có

a^3+b^3+3ab(a^2+b^2)+6ab(a+b)=a^3+b^3+3ab.a^2+3ab.b^2+6ab=a^3+b^3+3(a^2)b+3(b^2)a+3a(b-1)b^2+3b(a-1)a^2+6ab

                                               =(a+b)^3+3ab((b-1).b+(a-1).a)+6ab=(a+b)^3+3ab((1-b).(-b)+(1-a)(-a))+6ab=(a+b)^3+3ab(-2ab)+6ab

                                                                                                                                                        =(a+b)^3+(-6ab)ab+6ab

=>(a+b)^3+6ab(-ab-1)=6ab(-ab-1)+1 Vậy M=6ab(-ab-1)+1

k cho mình nhá

\(a^3+b^3+8=6ab\)

\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+8-6ab=0\)

\(\Leftrightarrow\left[\left(a+b\right)^3+2^3\right]-3ab\left(a+b+2\right)=0\)

\(\Leftrightarrow\left(a+b+2\right)\left[\left(a+b\right)^2+\left(a+b\right).2+4\right]-3ab\left(a+b+2\right)=0\)

\(\Leftrightarrow\left(a+b+2\right)\left(a^2+b^2+2ab+2a+2b+4-3ab\right)=0\)

\(\Leftrightarrow\left(a+b+2\right)\left(a^2+b^2-ab+2a+2b-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+2=0\\a^2+b^2-ab+2a+2b-4=0\end{matrix}\right.\)

\(\Leftrightarrow.....\)

Bài 2: 

a: Ta có: \(M=\left(x+y\right)^3+2x^2+4xy+2y^2\)

\(=\left(x+y\right)^3+2\cdot\left(x+y\right)^2\)

\(=7^3+2\cdot7^2=441\)

tks bn

Ta có: 

\(\left(3a-2b+c\right)^2=9a^2+4b^2+c^2+2\left(3ac-6ab-2bc\right)\)

\(\Rightarrow b^2=9a^2+4b^2+c^2\)

(vì \(3a-3b+c=0\Leftrightarrow3a-2b+c=-b\), \(6ab+2bc-3ac=0\))

\(\Leftrightarrow9a^2+3b^2+c^2=0\)

\(\Leftrightarrow a=b=c=0\). 

Khi đó: \(P=\left(-1\right)^{2019}+\left(-1\right)^{2020}+\left(-1\right)^{2021}=-1\)

Ta có: 

(3a−2b+c)2=9a2+4b2+c2+2(3ac−6ab−2bc)

⇒b2=9a2+4b2+c2

(vì 3a−3b+c=0⇔3a−2b+c=−b, 6ab+2bc−3ac=0)

⇔9a2+3b2+c2=0

⇔a=b=c=0. 

Khi đó: P=(−1)2019+(−1)2020+(−1)2021=−1