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(a+b+c)²=a²+b²+c²

Proved that 1/a+1/b+1/c=0

 Apr 25, 2015

Best Answer 

 #1
avatar+33586 
+5

(a+b+c)2 = a2+2*a*b+2*a*c+b2+2*b*c+c2

 

so we have 

a2+2*a*b+2*a*c+b2+2*b*c+c2 = a2+b2+c2

 

Subtract a2+b2+c2 from both sides

2*a*b+2*a*c+2*b*c = 0

 

Divide all terms by 2*a*b*c and we are left with

1/c + 1/b + 1/a = 0

 Apr 25, 2015
 #1
avatar+33586 
+5
Best Answer

(a+b+c)2 = a2+2*a*b+2*a*c+b2+2*b*c+c2

 

so we have 

a2+2*a*b+2*a*c+b2+2*b*c+c2 = a2+b2+c2

 

Subtract a2+b2+c2 from both sides

2*a*b+2*a*c+2*b*c = 0

 

Divide all terms by 2*a*b*c and we are left with

1/c + 1/b + 1/a = 0

Alan Apr 25, 2015

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