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If a + b + c = 56, a^2 + b^2 + c^2 = 1344, and a^2 = bc, then find abc.

 Dec 13, 2019
 #1
avatar+8875 
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If a + b + c = 56, a^2 + b^2 + c^2 = 1344, and a^2 = bc, then find abc.

 

Hello Guest!

 

\(a + b + c = 56\\ a^2 + b^2 + c^2 = 1344\\ a^2 = bc\\ b^2+c^2=1344-bc\\ (56-b-c)^2+b^2+c^2=1344\\ (b^2+2bc-112b+c^2-112c+3136)+b^2+c^2=1344\)

 

\(2b^2+2c^2-112b-112c+2bc=-1792\\ b^2+c^2-56b-56c+bc=-896\\ b^2+c^2=1344-bc\\ -56b-56c=-2240\\ b+c=40\\ a + b + c = 56\)

\(a=16 \)   

\(bc=256\\ b+c=40\\ (40-c)c=256\\ 40c-c^2=256\)

\(c^2-40c+256=0\\ c=20\pm\sqrt{400-256}\\ c_1=32\ entf\ddot{a}llt \\ c_2=8 \)

\(c=8\\ b=32\)

 

\(abc=4096\)

laugh  !

   

 Dec 13, 2019
 #2
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0

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

 Dec 13, 2019
 #3
avatar+106539 
+1

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

 

 

cool cool cool

 Dec 13, 2019

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