Problem
ALG-B1-M03-P011 Sum of Cubes
#11
★★☆☆☆ Level 2 of 5
Find \(x,y\) if \(x+y=6\), \(x^3+y^3=72\).
Express the sum of cubes through \(xy\).
\(x^3+y^3=(x+y)^3-3xy(x+y)=216-18xy\).
From \(216-18xy=72\), we get \(xy=8\). Then \(x,y\) are roots of \(t^2-6t+8=0\), so \((x,y)=(2,4)\), \((4,2)\).
A standard olympiad link: cubes plus Vieta.