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hartnn
 4 years 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 ??
hartnn
 4 years 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 ??

This Question is Closed

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1just for safety: a\(\ne\)b a\(\ne\)b

ParthKohli
 4 years ago
Best 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. :(

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1hmm....nice try no other formula for a^3+b^3 ?

nipunmalhotra93
 4 years ago
Best ResponseYou've already chosen the best response.3Are 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)

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1yes, question is correct.

nipunmalhotra93
 4 years ago
Best ResponseYou've already chosen the best response.3continuing from parthkohli, dw:1347876944383:dw

nipunmalhotra93
 4 years ago
Best ResponseYou've already chosen the best response.3and your question simplifies to dw:1347876996875:dw

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1yes, thats right, its 0.there is easier way. rather than continuing...

nipunmalhotra93
 4 years ago
Best ResponseYou've already chosen the best response.3@mukushla thanks :) @hartnn what is it?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1\((a^3+b^3)=(a+b)^33ab(a+b)=0\) and the rest is ............

nipunmalhotra93
 4 years ago
Best ResponseYou've already chosen the best response.3haha... yeah :D

nipunmalhotra93
 4 years ago
Best ResponseYou've already chosen the best response.3well even though we solved it, it IS technically wrong I think? :\

nipunmalhotra93
 4 years ago
Best ResponseYou've already chosen the best response.3dw:1347877692079:dw

nipunmalhotra93
 4 years ago
Best ResponseYou've already chosen the best response.3dw:1347877816341:dw

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

nipunmalhotra93
 4 years ago
Best ResponseYou've already chosen the best response.3Yea...to be clearer, I mean that a^2+b^2ab=0 iff a=b=0

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1u sure about a^2+b^2ab=0 iff a=b=0 ?

nipunmalhotra93
 4 years ago
Best ResponseYou've already chosen the best response.3yea... you can try that... you won't find any a, b s.t a,b are not both zero and the condition is satisfied....

nipunmalhotra93
 4 years ago
Best ResponseYou've already chosen the best response.3algebraically, x^3, is a one one function... so, a^3 +b^3=0 implies a^3=(b)^3 which implies a=b

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1a^3 =b^3 have 3 roots.....i think

nipunmalhotra93
 4 years ago
Best ResponseYou've already chosen the best response.3I 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.

nipunmalhotra93
 4 years ago
Best ResponseYou've already chosen the best response.3If your question covers complex numbers, then it's fine.... but for real, this is it :)

nipunmalhotra93
 4 years ago
Best ResponseYou've already chosen the best response.3@hartnn Thanks for the time man.. that was a good exercise for me ^_^

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

nipunmalhotra93
 4 years ago
Best ResponseYou've already chosen the best response.3you're most welcome buddy!
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