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If a³ + b³ = 0 , then the value of log (a + b) – 0.5(log a + log b + log 3) is equal to ??
 one year ago
 one year ago
Anyone getting bored? try this : If a³ + b³ = 0 , then the value of log (a + b) – 0.5(log a + log b + log 3) is equal to ??
 one year ago
 one year ago

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hartnnBest ResponseYou've already chosen the best response.1
just for safety: a\(\ne\)b a\(\ne\)b
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
\[a^3 + b^3 = (a + b)(a^2  ab + b^2)\]So,\[(a + b)(a^2  ab + b^2) = 0\]Dividing both sides by a + b,\[a^2  ab + b^ 2 = 0\]Since \(a^2  2ab + b^2 = (a  b)^2\), we have,\[a^2  2ab + b^2 +ab = 0\]\[(a  b)^2 + ab = 0\]\[(a  b)^2 =  ab\]\[a^2 + b^2 = ab\]One such case is a = 0 and b = 0, but you haven't allowed that. :(
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
hmm....nice try no other formula for a^3+b^3 ?
 one year ago

nipunmalhotra93Best ResponseYou've already chosen the best response.3
Are you sure that your question is correct? If a^3 + b^3=0, then either a=b or a=b=0.... :\ (because x^3 if a one one function)
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
yes, question is correct.
 one year ago

nipunmalhotra93Best ResponseYou've already chosen the best response.3
continuing from parthkohli, dw:1347876944383:dw
 one year ago

nipunmalhotra93Best ResponseYou've already chosen the best response.3
and your question simplifies to dw:1347876996875:dw
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
yes, thats right, its 0.there is easier way. rather than continuing...
 one year ago

nipunmalhotra93Best ResponseYou've already chosen the best response.3
@mukushla thanks :) @hartnn what is it?
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
\((a^3+b^3)=(a+b)^33ab(a+b)=0\) and the rest is ............
 one year ago

nipunmalhotra93Best ResponseYou've already chosen the best response.3
haha... yeah :D
 one year ago

nipunmalhotra93Best ResponseYou've already chosen the best response.3
well even though we solved it, it IS technically wrong I think? :\
 one year ago

nipunmalhotra93Best ResponseYou've already chosen the best response.3
dw:1347877692079:dw
 one year ago

nipunmalhotra93Best ResponseYou've already chosen the best response.3
dw:1347877816341:dw
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
if i read it correctly, there exist no a,b not equal to 0, such that a^2+b^2ab = 0.....??
 one year ago

nipunmalhotra93Best ResponseYou've already chosen the best response.3
Yea...to be clearer, I mean that a^2+b^2ab=0 iff a=b=0
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
u sure about a^2+b^2ab=0 iff a=b=0 ?
 one year ago

nipunmalhotra93Best ResponseYou've already chosen the best response.3
yea... you can try that... you won't find any a, b s.t a,b are not both zero and the condition is satisfied....
 one year ago

nipunmalhotra93Best ResponseYou've already chosen the best response.3
algebraically, x^3, is a one one function... so, a^3 +b^3=0 implies a^3=(b)^3 which implies a=b
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
a^3 =b^3 have 3 roots.....i think
 one year ago

nipunmalhotra93Best ResponseYou've already chosen the best response.3
I plotted two graphs.... z=xy and z=x^2+y^2 they intersect only at (0,0) Yes there are three values of b for every a. But they are complex.
 one year ago

nipunmalhotra93Best ResponseYou've already chosen the best response.3
If your question covers complex numbers, then it's fine.... but for real, this is it :)
 one year ago

nipunmalhotra93Best ResponseYou've already chosen the best response.3
@hartnn Thanks for the time man.. that was a good exercise for me ^_^
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
hmm..log of complex numbers....didn't think of it b4 posting....and thanks to u for wonderful explanation.
 one year ago

nipunmalhotra93Best ResponseYou've already chosen the best response.3
you're most welcome buddy!
 one year ago
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