Note that a−1=yx and b−1=xy are reciprocals. That is, (a−1)(b−1)=1⟹ab−a−b+1=1⟹ab=a+b Let t=ab=a+b. Then we can write a2+b2=(a+b)2−2ab=t2−2t so t2−2t=15, which factors as (t−5)(t+3)=0. Since a,b>0, we must have t=5. Then, we compute a3+b3=(a+b)3−3ab(a+b)=53−3⋅52=50
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