Question

a cube+ b cube= 0 then log(a+ b)-1/2(log a+ log b+ log3)​

Answer 1

Question:-

a ³+ b ³= 0 then log(a+ b)-1/2(log a+ log b+ log3)

Answer:-

A³ +B³ =(A+B)(A² + B² -AB)=0.

as (A+B) can’t be zero, (A² + B² -AB)=0.

A² + B²=AB. …….1)

log(A+B)=0.5log((A+B)^2)=0.5log(A^2 + B^2 + 2AB)…..2)

FROM 1) AND 2)

Log(A+B)=0.5Log(3AB)…..3)

now,

logA + logB + log3=log(3AB)…..4)

using 3 and 4 you get your answer as zero.

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Answer 2

Hey Buddy

Here's The Answer

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Given :

a³ + b³ = 0

To find :

log ( a+b ) - 1/2 ( log a + log b + log 3 )

We know

=> log a + log b = log ( a.b)

=> log a - log b = log ( a/b )

Now

=> log ( a+b ) - 1/2 ( log ( a.b ) + log 3 )

=> log ( a+b ) - 1/2 ( log ( 3.a.b ) )

=> log ( a+b ) - log ( 3ab )¹´²

=> log ( a+b / √3ab ) ________ ( 1 )

Now

a³ + b³ = 0

=> ( a² + b² - ab ) ( a + b ) = 0

Now we have 2 cases

=> ( a + b ) = 0

If we put ( a + b = 0 ) in eq ( 1 ), the log will become 0, so it's can't be done.

=> ( a² + b² - ab ) = 0

=> a² + b² + 2ab - 2ab - ab = 0 ( modification )

=> ( a + b )² - 3ab = 0

=> ( a + b )² = 3ab

=> a + b = ( 3ab )¹´²

=> a + b = √ 3ab

Now putting value of ( a + b ) = √ 3ab in eq ( 1 )

=> log ( a+b / √3ab )

=> log ( √3ab / √3ab )

=> log 1

=> 0 ✓

Hope It Helps.