Problem: Three times Dick's age plus Tom's age equals twice Harry's age. Double the cube of Harry's age is equal to three times the cube of Dick's age added to the cube of Tom's age. Their respective ages are relatively prime to each other. The sum of the squares of their ages is
Answer Choices:
A. 42
B. 46
C. 122
D. 290
E. 326
Solution:
Let T,D and H denote the ages of Tom, Dick, and Harry respectively. Then 3D+T=2H and 2H3=3D3+T3 or equivalently 2(H−D)=D+T and 2(H3−D3)=D3+T3. Since, D+Tî€ =0,2(H−D)2(H3−D3)​=D+TD2+T3​ or H2+HD+B2=B2−DT+T2. Hence T2−H2=D(T+H) so that T−H=D. Eliminating T from the last and very first equation gives H=4D so that D=1 and H=4 because H and D are relatively prime integers. Also T=H+D=5 and the required sum of squares is T2+D2+H2=52+12+42=42.