Từ \(x+y=a+b\Rightarrow\left(x+y\right)^2=\left(a+b\right)^2\)
\(\Rightarrow x^2+2xy+y^2=a^2+2ab+b^2\)
Do \(x^2+y^2=a^2+b^2\Rightarrow2xy=2ab\Rightarrow xy=ab\)
\(\Rightarrow-xy=-ab\)
Ta có: \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=\left(a+b\right)\left(a^2-ab+b^2\right)\)\(=a^3+b^3\)
Hay \(x^3+y^3=a^3+b^3\left(đpcm\right)\)
Theo bài ra ta có :
\(x+y=a+b\\ \Rightarrow\left(x+y\right)^2=\left(a+b\right)^2\\ \Rightarrow x^2+2xy+y^2=a^2+2ab+b^2\\ \text{Mà }x^2+y^2\\ =a^2+b^2\\ \Rightarrow2xy=2ab\\ \Rightarrow xy=ab\\ \Rightarrow\left(x^2+y^2\right)-xy=\left(a^2+b^2\right)-ab\\ \Rightarrow x^2-xy+y^2= a^2-ab+b^2\\ \Rightarrow\left(x+y\right)\left(x^2-xy+y^2\right)=\left(a+b\right)\left(a^2-ab+b^2\right)\\ \Rightarrow x^3+y^3=a^3+b^3\left(đpcm\right)\)
Vậy...........................................................(ghi lại đpcm)