a) ta có: \(\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3\)(1)

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

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

từ (1) và (2) \(\Rightarrow\left(a-b\right)^3=-\left(b-a\right)^3\)

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

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

                                     \(=a^2+2ab+b^2\)(4)

từ (3) và (4) \(\Rightarrow\left(-a-b\right)^2=\left(a+b\right)^2\)

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

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

Học tốt!

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

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

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

a,(a-b)³=a³-3a²b+3ab²-b³=-(b³-3ab²+3a²b-a³)=-(b-a)³

b,(-a-b)²=(-1)².(a+b)²=(a+b)²

(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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

a) Sử dụng tính chất hai số đối nhau:

(a – b)³ = [(–1)(b – a)]³ =(–1)³(b – a)³ = –1.(b – a)³ = –(b – a)³   (đpcm)

b) (–a – b)² = [(– 1).(a + b)]² = (–1)²(a + b)² = 1.(a + b)² = (a + b)² (đpcm)

Học tốt !

a, ( a - b )³ 

= [( - 1 ) ( b - a )]³

=( - 1 )³ ( b - a ) ³

= - 1 . ( b - a ) ³

= - ( b - a ) ³  ( đpcm )

b , ( - a - b ) ²

= [ ( - 1 ) . ( a + b ) ]

= ( - 1 ) ² ( a + b ) ²

= 1 . ( a + b ) ²

= ( a + b ) 2 ( đpcm )

(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

help meeeeeeeeeeeeeeeeeeeeeeeeeeeeee

1) a³+b³+c³-3abc = (a+b)³-3ab(a+b)+c³-3abc

                           = (a+b+c)(a²+2ab+b²-ab-ac+c²) -3ab(a+b+c)

                           = (a+b+c)( a²+b²+c²-ab-bc-ca)

Vì a+b+c=0

=> a+b=-c

=> (a+b)³= (-c)³

=> a³+b³+3ab(a+b) = (-c)³

=> a³+b³+c³= 3abc