\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=a.\left(b-xy\right)\left(1\right)\)
Ta có :
\(\left(x+y\right)^2=a^2=x^2+2xy+y^2=b+2xy\)
\(=>2xy=a^2-b\)
\(=>xy=\dfrac{a^2-b}{2}\)
Thay vào (1) ta có ĐPCM
CHÚC BẠN HỌC TỐT....
Có .\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\) Mà x+y=a;\(x^2+y^2=b\Rightarrow x^2-xy+y^2=b-xy\). Nên \(x^3+y^3=a\left(b-xy\right)\)
Ta có : x + y = a
=> ( x + y )2 = a2
=> x2 + 2xy + y2 = a2
=> 2xy = a2 - ( x2 + y2 )
=> xy = \(\dfrac{a^2-\left(x^2+y^2\right)}{2}\)
=> xy = \(\dfrac{a^2-b}{2}\)
Lại có : x3 + y3 = ( x + y ) . ( x2 - xy + y2 )
=> x3 + y3 = a .( b - \(\dfrac{a^2-b}{2}\) )
Ta có:
\(x^2+y^2=\left(x+y\right)^2-2xy\)
\(\Rightarrow b=a^2-2xy\)
\(\Rightarrow2xy=a^2-b\)
\(\Rightarrow xy=\dfrac{a^2-b}{2}\left(1\right)\)
Ta lại có:
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\)
\(\Rightarrow x^3+y^3=a^3-3xy.a\)
Thay (1) vào biểu thức trên ta có:
\(x^3+y^3=a^3-3.\dfrac{a^2-b}{2}.a\)
\(x^3+y^3=a^3-\dfrac{\left(3a^2-3b\right).a}{2}\)
\(x^3+y^3=a^3-\dfrac{3a^3-3ab}{2}\)
\(x^3+y^3=\dfrac{2a^3-3a^3+3ab}{2}\)
\(x^3+y^3=\dfrac{-a^3+3ab}{2}\)
