(a-b)³=a³-3a²b+3ab²-b³ (1)

-(b-a)³=-(b³-3b²a+3ba²-a³)=-b³+3ab²-3a²b+a³=a³-3a²b+3ab²-b³ (2)

từ (1) và (2) => VT=VP => đpcm.

(-a-b)²=[(-a)+(-b)]²=(-a)²+2.(-a).(-b)+(-b)²=a²+2ab+b²=(a+b)²

=> VT=VP => đpcm.

a/ -(b-a)^3= -(b^3-3b^2a+3ba^2-a^3)

              = -b^3+3ab^2a-3ba^2+a^3

             = (a-b)^3

b/ tương tự ta dùng hằng đẳng thức để chứng minh

a) a - b = - (b - a) = (-1)*(b - a)

=> (a - b)³ = [(-1)*(b - a)]³ = (-1)³ * (b - a)³ = -(b - a)³

b) -(a + b) = (- a - b)

=> (-1)² * (a + b)² = (-a - b)²

=> (-a -b)² = (a + b)²

a) (a-b)^3=-(b-a)^3

\(Taco:-\left(b-a\right)^3\)

=\(-\left(b-a\right)\left(b-a\right)\left(b-a\right)\)

\(=\left(a-b\right)\left(b-a\right)\left(b-a\right)\)

\(=-\left(a-b\right)\left(a-b\right)\left(b-a\right)\)

\(=\left(a-b\right)\left(a-b\right)\left(a-b\right)=\left(a-b\right)^3\)

\(\left(-a-b\right)^2=\left(-a-b\right)\left(-a-b\right)\)

\(=-\left(a+b\right)\left(-a-b\right)\)

\(=\left(a+b\right)\left(a+b\right)\)

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

\(a,\left(a-b\right)^3=-\left(b-a\right)^3\)

\(=-\left(b-a\right)\left(b-a\right)\left(b-a\right)\)

\(=\left(a-b\right)\left(b-a\right)\left(b-a\right)\)

\(=-\left(a-b\right)\left(a-b\right)\left(b-a\right)\)

\(=\left(a-b\right)\left(a-b\right)\left(a-b\right)\)

\(=\left(a-b\right)^3\)

\(b,\left(-a-b\right)^2=\left(-a-b\right)\left(-a-b\right)\)

\(=-\left(a+b\right)\left(-a-b\right)\)

\(=\left(a+b\right)\left(a+b\right)\)

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

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

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

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

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

\(=VP\left(đpcm\right)\)         

Ta có: \(a^3-b^3+ab\left(a-b\right)=\left(a-b\right)\left(a^2+ab+b^2\right)+ab\left(a-b\right)\)

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

\(=\left(a-b\right)\left(a+b\right)^2\)( đpcm )

Ta có:

VT=a³-b³+ab(a-b)

VT=a³-b³+a²b-ab²

VT=(a³+a²b)-(b³+ab²)

VT=a²(a+b)-b²(b+a)

VT=(a+b)(a²-b²)

VT=(a+b)(a-b)(a+b)

VT=(a-b)(a+b)²=VP

a) Biến đổi VT . Mẫu chung là ( a + 2b )( a - 2b )

\(VT=\frac{a+2b-6b-2\left(a-2b\right)}{a^2-4b^2}=-\frac{a}{a^2-4b^2}\)( 1 )

Biến đổi VP 

\(-\frac{1}{2a}\left(\frac{a^2+4b^2}{a^2-4b^2}+1\right)=-\frac{1}{2a}\cdot\frac{a^2+4b^2+a^2-4b^2}{a^2-4b^2}\)

\(=-\frac{1}{2a}\cdot\frac{2a^2}{a^2-4b^2}=-\frac{a}{a^2-4b^2}\)( 2 )

Từ ( 1 ) và ( 2 ) => VT = VP ( đpcm )

b) \(a^3+b^3+\left(\frac{b\left(2a^3+b^3\right)}{a^3-b^3}\right)=\left(\frac{a\left(a^3+2b^3\right)}{a^3-b^3}\right)^3\)

<=> \(b^3+\left(\frac{b\left(2a^3+b^3\right)}{a^3-b^3}\right)^3=\left(\frac{a\left(a^3+2b^3\right)}{a^3-b^3}\right)-a^3\)( * )

Biến đổi VT của ( * ) ta có :

\(VT=\left[b+\frac{b\left(2a^3+b^3\right)}{a^3-b^3}\right]\left[b^2-\frac{b^2\left(2a^3+b^3\right)}{a^3-b^3}+\frac{b^2\left(2a^3+b^3\right)^2}{\left(a^3-b^3\right)^2}\right]\)

\(=\frac{3a^3b}{a^3-b^3}\cdot\frac{3a^6b^2+3a^3b^5+3b^8}{\left(a^3-b^3\right)^2}\)

\(=\frac{9a^3b^3}{\left(a^3-b^3\right)^3}\left(a^6+a^3b^3+b^6\right)\)( 1 )

\(VP=\left[\frac{a\left(a^3+2b^3\right)}{a^3-b^3}-a\right]\left[\frac{a^2\left(a^3+2b^3\right)^2}{\left(a^3-b^3\right)^2}+\frac{a^2\left(a^3+2b^3\right)}{a^3-b^3}+a^2\right]\)

\(=\frac{3ab^3}{a^3-b^3}\cdot\frac{3a^8+3a^5b^3+3a^2b^6}{\left(a^3-b^3\right)^2}\)

\(=\frac{9a^3b^3}{\left(a^3-b^3\right)^3}\left(a^6+a^3b^3+b^6\right)\)( 2 )

Từ ( 1 ) và ( 2 ) => VT = VP => ( * ) đúng 

=> Hằng đẳng thức đúng 

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

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

(a+3)^3=(a+b)^2*(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

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

Không có được CM ngược lại bạn ạ

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