Three innocent fractions
If a + b + c = 0 , but a b c = 0 , then
b c a 2 + a c b 2 + a b c 2 = ?
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7 solutions
Brilliant thinking!
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There's nothing brilliant in that.Even I got the same answer in the same method.
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I think he meant the elegance of the solution, and if you had gotten the answer than why didn't you post a solution?
a+b+c=0 so a+b=-c cubing on both sides we get...a^3+b^3+3ab(a+b)=-(c^3) rearranging,..a^3+b^3+c^3= -3ab(a+b) but we know a+b=-c a^3+b^3+c^3= 3abc…(1)
and, (a^2÷bc) + (b^2÷ac) + (c^2÷ab) =(a^3+b^3+c^3)÷abc……(2) substituting (1) in (2)... we get (3abc)÷(abc)=3
Can someone explain why a^3+b^3+c^3=3abc?
Same method :)
one question from work and energy .. When you lift a box from the floor and put it on an almirah the potential energy of the box increases ,but there is no change in its kinetic energy. Is it a violation of conservation of energy?
Because the variables are not equal to zero. a = − ( b + c ) , b = − ( a + c ) c = − ( a + b ) .
Therefore b c b 2 + 2 b c + c 2 + a c a 2 + 2 a c + c 2 + a b a 2 + 2 a b + b 2 . Simplifying c b + 2 + b c + c a + 2 + a c + b a + 2 + a b = 6 + c b + a + b c + a + a c + b . Combining like terms 6 + − ( b + a ) b + a + − ( c + a ) c + a + − ( c + b ) c + b = 6 + − 3 = 3 .
I think the same :3
Given: a+b+c=0 and abc ≠0 a+b+c=0 implies a=-b-c, b=-a-c, c=-a-b
a2/bc = [(-1)2(b+c)2]/bc = [b2+2bc+c2]/bc b2/ac = [(-1)2(a+c)2]/ac = [a2+2ac+c2]/ac c2/ab = [(-1)2(a+b)2]/ab = [a2+2ab+b2]/ab
a2/bc + b2/ac + c2/ab = [b2+2bc+c2]/bc + [a2+2ac+c2]/ac + [a2+2ab+b2]/ab = [a/a] [b2+2bc+c2]/bc + [b/b] [a2+2ac+c2]/ac + [c/c] [a2+2ab+b2]/ab = [ab2 + 2abc + ac2 + ba2 + 2abc + bc2 + ca2 + 2abc +cb2 ]/abc = b/c + c/b +a/c + c/a + a/c + a/b + b/a + 2 + 2 + 2 = [a+c]/b + [b+c]/a + [a +b]/c + 6
Substituting a=-b-c, b=-a-c, c=-a-b = -b/b + -a/a + -c/c + 6 = -1 + -1 + -1 + 6 = 3
Oh! I kept trying to do this method, but never realized the substitution at the end.
there was hardly a need to do so....it was very easy
I didn't solve at all but simply I just picked the lowest possible numbers that would eaual to 0 which is: a+b+c=0 2+(-1)+(-1)=0 2+(-2)=0 2-2=0 0=0
Then, abc != 0 2(-1)(-1) != 0 2(1) != 0 2 != 0
So then I substituted a = 2, b = -1, c =-1
You got lucky, this doesn't prove the general answer.
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I wouldn't call it lucky, if you plug any three (different) numbers that add up to zero, the result will be three. You can test it with any three (different) numbers that add up to zero. However, it is also true that proving the general case is more difficult. Still, I wouldn't call testing the specific case lucky. Just not as impressive.
@Jerry Manubay You just proved that it was true for specific values of a,b,c.You have to prove that it's true for all values
My simple approach. Just assume three
numbers when added is equals to zero.
Say a=2, b=2 and c=-4
4/-8 +4/-8 +16/4=-24/-8
=3
Let a=1, b=2, c=-3 Then a+b+c=1+2-3=0 Also, abc=(1)(2)(-3)=-6 (not equal to 0) Now, a^2=1, b^2=4, c^2=9 Also, bc=-6, ca=-3, ab=2 Then (-1/6)+(-4/3)+(9/2)=(-1-8+27)/6=18/6=3.
Bad solutions people..!! Better use AM GM inequality will be best and easy solutions..!! no need expansion..! :)
Please show how you would do it using the AM GM inequality.
AM-GM is for non-negative real numbers only. If a , b , c = 0 and a + b + c = 0 , then at least one must be negative, so AM-GM doesn't apply.
Note that a 3 + b 3 + c 3 = ( a + b + c ) ( a 2 + b 2 + c 2 − a b − b c − c a ) + 3 a b c , so when a + b + c = 0 , then a 3 + b 3 + c 3 = 3 a b c .
Making the common denominator of a b c , we have a b c a 3 + a b c b 3 + a b c c 3 = a b c a 3 + b 3 + c 3 = a b c 3 a b c = 3 .