If a + b + c = 56, a^2 + b^2 + c^2 = 1344, and a^2 = bc, then find abc.
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\(a + b + c = 56\\ a^2 + b^2 + c^2 = 1344\\ a^2 = bc\\ b^2+c^2=1344-bc\\ (56-b-c)^2+b^2+c^2=1344\\ (b^2+2bc-112b+c^2-112c+3136)+b^2+c^2=1344\)
\(2b^2+2c^2-112b-112c+2bc=-1792\\ b^2+c^2-56b-56c+bc=-896\\ b^2+c^2=1344-bc\\ -56b-56c=-2240\\ b+c=40\\ a + b + c = 56\)
\(a=16 \)
\(bc=256\\ b+c=40\\ (40-c)c=256\\ 40c-c^2=256\)
\(c^2-40c+256=0\\ c=20\pm\sqrt{400-256}\\ c_1=32\ entf\ddot{a}llt \\ c_2=8 \)
\(c=8\\ b=32\)
\(abc=4096\)
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a=1; b=1;c=1;d=a + b + c;e=a^2+b^2+c^2;f=a^2 ;if(d==56 and e==1344 and f==c*b, goto7, goto8);printa, b, c; a++;if(a<100, goto3, 0);a=1;b++;if(b<100, goto3, discard=0; a=1;b=2;c++;if(c<100, goto3, 0)
a b c
16 32 8
16 8 32
a + b + c = 56 square both sides
a^2 + b^2 + c^2 + 2(ab + ac + bc) = 3136 (1)
And
a^2 + b^2 + c^2 = 1344 (2)
Sub (2) into (1)
1344 + 2(ab + ac + bc) = 3136 subtract 1344 from both sides
2(ab + ac + bc) = 1792 divide both sides by 2
(ab + ac + bc) = 896 (a^2 =bc )
(ab + ac + a^2) = 896 factor out a
a( a + b + c) = 896
a ( 56) = 896 divide both sides by 56
a = 16
And a^2 = bc
So
abc = a *a^2 = a^3 =16^3 = 4096