x3+y3=(x+y)(x2-xy+y2)=(x+y(x2+2xy+y2-3xy)=(x+y)[(x+y)2-3xy]=4[16-6]=40
Đáp số: 40
\(x+y=4\Rightarrow\left(x+y\right)^2=4^2\Leftrightarrow x^2+2xy+y^2=16\Leftrightarrow x^2+2.2+y^2=16\)
\(\Leftrightarrow x^2+4+y^2=16\Leftrightarrow x^2+y^2=12\)
=>\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=4\left(12-2\right)=4.10=40\)
\(\left(x+y\right)^3=x^3+y^3+3xy\left(x+y\right)\)
=\(x^3+y^3+24=64=>x^3+y^3=40\)
\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)=\(\left(x+y\right)\left(x^2+2.xy-3.xy+y^2\right)\)=\(\left(x+y\right)\left(\left(x+y\right)^2-3.xy\right)\)
Thay vô = 4.(16-3.2)=40
k đúng nhé bạn
