\(x+y=a\left(1\right)\)
\(x-y=b\left(2\right)\)
\(\left(1\right)+\left(2\right)\Rightarrow2x=a+b\Rightarrow x=\dfrac{a+b}{2}\)
\(\left(1\right)\Rightarrow y=a-x\Rightarrow y=a-\dfrac{a+b}{2}\Rightarrow y=\dfrac{a-b}{2}\)
\(xy=\dfrac{\left(a+b\right)}{2}.\dfrac{\left(a-b\right)}{2}=\dfrac{a^2-b^2}{4}\)
\(x^3-y^3=\left(\dfrac{a+b}{2}\right)^3-\left(\dfrac{a-b}{2}\right)^3=\dfrac{\left(a+b\right)^3}{8}-\dfrac{\left(a-b\right)^3}{8}\)
\(=\dfrac{\left(a+b\right)^3-\left(a-b\right)^3}{8}\)
\(=\dfrac{\left(a+b-a+b\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]}{8}\)
\(=\dfrac{2b\left[a^2+b^2+2ab+a^2-b^2+a^2+b^2-2ab\right]}{8}\)
\(=\dfrac{b\left[3a^2+b^2+2ab\right]}{4}\)
\(\left\{{}\begin{matrix}x+y=a\\x-y=b\end{matrix}\right.\) tính \(x^3\) - y3 theo \(a\) và \(b\)
⇒ \(\left\{{}\begin{matrix}x+y+x-y=a+b\\x-y=b\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}2x=a+b\\y=x-b\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}x=\left(a+b\right):2\\y=\left(a-b\right):2\end{matrix}\right.\) ⇒ \(xy\) = \(\dfrac{a+b}{2}\)\(\times\)\(\dfrac{a-b}{2}\) = \(\dfrac{a^2-b^2}{4}\)
\(x^{3^{ }}\) - y3 = (\(x\) - y)(\(x^2\) + \(x\)y + y2) = \(\left(x-y\right)\)\(\left(\left[x+y\right]^2-xy\right)\) (1)
Thay \(x-y\) = a; \(x\) + y = b và \(xy\) = \(\dfrac{a^2-b^2}{4}\) vào (1) ta có:
\(x^3\) - y3 = b.(a2 - \(\dfrac{a^2-b^2}{4}\)) = b.\(\dfrac{3a^2+b^2}{4}\) = \(\dfrac{3a^2b+b^3}{4}\)
