Question

if a^1/x=b^2/y=c^3/z and abc =1, prove that x+y+z=0​

Answer 1

Your question needs a correction.

To prove : 6x + 3y + 2z = 0

Answer:

0 = x + y / 2 + z / 3

Or, 0 = 6x + 3y + 2z

Step-by-step explanation:

Given,

• a^( 1 / x ) = b^( 2 / y ) = c^( 3 / z )
• abc = 1

Let,

a^( 1 / x ) = b^( 2 / y ) = c^( 3 / z ) = k

Case 1 : a^( 1 / x ) = k

= > a^( 1 / x ) = k

= > { a^( 1 / x ) }^x = k^x { if a = b, then a^n = b^n }

= > a^( x × 1 / x ) = k^x

= > a = k^x ...( 1 )

Case 2 : b^( 2 / y ) = k

= > b^( 2 / y ) = k

= > { b^( 2 / y ) }^y = k^y

= > b^( 2 / y × y ) = k^y

= > b^2 = k^y

= > b = √( k^y ) or k^( y / 2 ) ...( 2 )

Case 3 : c^( 3 / z ) = k

= > c^( 3 / z ) = k

= > { c^( 3 / z ) }^z = k^z

= > c^3 = k^z

= > c = $\sqrt[3]{k^z}$ or k^( z / 3 ) ...( 3 )

Multiplying ( 1 ), ( 2 ) and ( 3 ) :

= > a × b × c = k^x × k^( y / 2 ) x k^( z / 3 )

= > abc = k^( x + y / 2 + z / 3 )

= > 1 = k^( x + y / 2 + z / 3 ) { given, abc = 1 }

= > k^0 = k^( x + y / 2 + z / 3 ) { any number having 0 as it's powerful is equal to 1 , k^0 = 1 , where k ≠ 0 }

= > 0 = x + y / 2 + z / 3

Or, 0 = 6x + 3y + 2z

Hence proved.