If \(x + y = 12\) and \(x - y = 4\), find \(\dfrac{x^{3}+y^{3}}{x+y}\).
48
Given: \(x + y = 12\) and \(x - y = 4\).
Solve the two linear equations. Adding them: \(2x=16\Rightarrow x=8\).
Subtracting: \(2y=8\Rightarrow y=4\).
Use \(x^{3}+y^{3}=(x+y)(x^{2}-xy+y^{2})\), so \(\dfrac{x^{3}+y^{3}}{x+y}=x^{2}-xy+y^{2}\).
Compute \(x^{2}=64\), \(y^{2}=16\), \(xy=32\).
\(x^{2}-xy+y^{2}=64-32+16\).
\(=48\).
Hence the value of \(\dfrac{x^{3}+y^{3}}{x+y}\) is 48.
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