Có : a³ - b³ - 84

= (a - b)(a² + ab + b²) - 84

= 6.(a² + b² + 9) - 84

= 6a² + 6b² + 54 - 84

= 6(a² + b²) - 30

= 6 [ (a - b)² + 2ab ] - 30

= 6 ( 6² + 2.9 ) - 30

= 324 - 30

= 294

 a³ - b³ - 84

= (a - b)(a² + ab + b²) - 84

= 6.(a² + b² + 9) - 84

= 6a² + 6b² + 54 - 84

= 6(a² + b²) - 30

= 6 [ (a - b)² + 2ab ] - 30

= 6 ( 6² + 2.9 ) - 30

= 324 - 30

= 294

\(A=a^3-b^3-84\)

\(=\left(a-b\right)\left(a^2+ab+b^2\right)-84\)

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

\(=6.\left[6^2+3.9\right]=6.63=379\)

\(Ủng\)hộ nhak

Theo bài ra ta có : 

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

Thay \(a-b=6;a.b=9\)vào biểu thức trên ta , có : 

\(6^3+3.9.6=378\)

mà biểu thức - 84 

Vậy \(378-84=294\)

Vậy \(a^3-b^3-84=294\)

ta có : a-b=6

nên (a-b)^2=36

a^2-2ab+b^2=36

mà ab=9

=>a^2+b^2-2.9=36

a^2+b^2=54

lại có A=a^3-b^3-84

=(a-b)(a^2+ab+b^2)-84

=6.(a^2+b^2+9)-84

=6.(54+9)-84

=378-84=294

Ta có : \(a-b=6\Rightarrow\left(a-b\right)^2=36\Rightarrow a^2-2ab+b^2=36\Rightarrow a^2+b^2=54\)

\(A=a^3-b^3-84=\left(a-b\right)\left(a^2+b^2+ab\right)-84=6.\left(54+9\right)-84=294\)

A=a³-b³-84 =(a-b)(a²+ab+b²)-84=(a-b){(a-b)²+3ab}=6.[6²+3.9]=6.63=378

\(A=a^3-b^3-84=\left(a-b\right)^3+3xy\left(x-y\right)-84=6^3+3.9.6-84=-30\)

a)\(a+b=-5\)

\(\Rightarrow\left(a+b\right)^2=25\)

\(\Leftrightarrow a^2+2ab+b^2=25\)

\(\Leftrightarrow a^2+12+b^2=25\)

\(\Leftrightarrow a^2+b^2=13\)

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

\(=-5\left(13-6\right)=-35\)

b) \(a-b=9\)

\(\Leftrightarrow\left(a-b\right)^2=81\)

\(\Leftrightarrow a^2-2ab+b^2=81\)

\(\Leftrightarrow a^2-44+b^2=81\)

\(\Leftrightarrow a^2+b^2=125\)

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

\(=9\left(125+22\right)=1323\)

a) Ta có: \(a^3+b^3\)

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

Thay \(ab=40\) và \(a+b=-6\) vào biểu thức ta có

\(\left(-6\right)^3-3\cdot7\cdot\left(-6\right)=-90\)

b) Ta có: \(a^3-b^3\)

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

Thay \(ab=40\) và \(a-b=3\) vào biểu thức ta có:

\(3^3+3\cdot40\cdot3=387\)

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

=(-6)^3-3*7*(-6)

=-90

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

=3^3+3*40*3

=387

\(P=a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)=\left(-1\right)^3-3\left(-6\right)\left(-1\right)=-1-18=-19\)

\(a+b=-1\Rightarrow\left(a+b\right)^2=1\)

\(\Rightarrow a^2+2ab+b^2=1\Rightarrow a^2+b^2=1-2ab=1-2.\left(-6\right)=13\)

\(P=a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)=\left(-1\right).\left(13+6\right)=-19\)

( a - b)^2 = 1 ^2 

=> a^2 - 2ab + b^2 = 1 

=> a^2 - 2.6 + b^2 = 1 

=> a^2 + b^2 = 1 + 12 = 13 

a^3 - b^3 = ( a- b)(a^2 + ab+ b^2) = 1 ( 13 + 6 ) = 19