SOLUTION: There are 3 different triples of(not necessarily distinct) positive integers(a,b,c, d,e,f and g,h,i) such that {{{ a^2+b^2+c^2=d^2+e^2+f^2=g^2+h^2+i^2=89 }}}. Evaluate the expressi

Algebra ->  Expressions -> SOLUTION: There are 3 different triples of(not necessarily distinct) positive integers(a,b,c, d,e,f and g,h,i) such that {{{ a^2+b^2+c^2=d^2+e^2+f^2=g^2+h^2+i^2=89 }}}. Evaluate the expressi      Log On


   

Question 1189715: There are 3 different triples of(not necessarily distinct) positive integers(a,b,c, d,e,f and g,h,i) such that +a%5E2%2Bb%5E2%2Bc%5E2=d%5E2%2Be%5E2%2Bf%5E2=g%5E2%2Bh%5E2%2Bi%5E2=89+. Evaluate the expression +%28a%2Bb%2Bc%29%28d%2Be%2Bf%29%28g%2Bh%2Bi%29+
Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
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The three possible ways to present the number 89 as the sum of three squares are


    89 =  9 + 64 + 16 = 3^2 + 8^2 + 4^2   ( so, (a,b,c) = (3,8,4) ),

    89 = 49 + 36 +  4 = 7^2 + 6^2 + 2^2   ( so, (d,e,f) = (7,6,2) ),

    89 = 81 +  4 +  4 = 9^2 + 2^2 + 2^2   ( so, (g,h,i) = (9,2,2) ).


Then  (a+b+c)*(d+e+f)*(g+h+i) = (3+8+4)*(7+6+2)*(9+2+2) = 15*15*13 = 2925.    ANSWER

Solved.