It seems so easy

Algebra Level 4

if x a = y b = z c \frac { x }{ a } =\frac { y }{ b } =\frac { z }{ c } then [ a 2 x 2 + b 2 y 2 + c 2 z 2 a 3 x + b 3 y + c 3 z ] 3 2 \quad { \left[ \frac { { a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+{ c }^{ 2 }{ z }^{ 2 } }{ { a }^{ 3 }x+{ b }^{ 3 }y+{ c }^{ 3 }z } \right] }^{ \frac { 3 }{ 2 } } =

(the problem is not original)

  1. x y z a b c \frac { xyz }{ abc }
  1. ( x y z ) 2 a b c \frac { ({ xyz) }^{ 2 } }{ \sqrt { abc } }
  1. x y z a b c \frac { \sqrt { xyz } }{ abc }
  1. x y z a b c \frac { \sqrt { xyz } }{ \sqrt { abc } }

Rishabh Jain by Rishabh Jain , 23, India

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2 solutions

Shivamani Patil
Jul 29, 2014

L e t x a = y b = z c = k T h e r e f o r e x = a k , y = b k , z = c k w e h a v e ( a 2 x 2 + b 2 y 2 + c 2 z 2 a 3 x + b 3 x + c 3 x ) 3 2 − − − − − − ( 1 ) s u b s t i t u t i n g v a l u e s ( 1 ) b e c o m e s ( a 4 k 2 + b 4 k 2 + c 4 k 2 a 4 k + b 4 k + c 4 k ) 3 2 = ( k 2 k ( a 4 + b 4 + c 4 a 4 + b 4 + c 4 ) ) 3 2 = k 3 2 w e k n o w t h a t k 3 = x y z a b c T h e r e f o r e k 3 2 = x y z a b c Let\quad \frac { x }{ a } =\frac { y }{ b } =\frac { z }{ c } =k\\ \\ Therefore\quad x=ak\quad ,\quad y=bk\quad ,\quad z=ck\\ we\quad have{ \left( \frac { { a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+{ c }^{ 2 }{ z }^{ 2 } }{ { a }^{ 3 }{ x }+{ b }^{ 3 }x+{ c }^{ 3 }x } \right) }^{ \frac { 3 }{ 2 } }------(1)\\ substituting\quad values\quad (1)\quad becomes\\ { \left( \frac { { a }^{ 4 }{ k }^{ 2 }+{ b }^{ 4 }{ k }^{ 2 }+{ c }^{ 4 }{ k }^{ 2 } }{ { a }^{ 4 }{ k }+{ b }^{ 4 }k+{ c }^{ 4 }k } \right) }^{ \frac { 3 }{ 2 } }\\ ={ \left( \frac { { k }^{ 2 } }{ k } { \left( \frac { { a }^{ 4 }+{ b }^{ 4 }+{ c }^{ 4 } }{ { a }^{ 4 }+{ b }^{ 4 }+{ c }^{ 4 } } \right) } \right) }^{ \frac { 3 }{ 2 } }\\ ={ k }^{ \frac { 3 }{ 2 } }\\ we\quad know\quad that\quad \\ { k }^{ 3 }=\frac { xyz }{ abc } \\ Therefore\quad \\ { k }^{ \frac { 3 }{ 2 } }=\sqrt { \frac { xyz }{ abc } } \\ \\ \\ \\ \\

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice Solution! But for those who want the answer in less than one second, observe the options carefully. If you are creative, you'll see that the options are numbered, and by default, you get option one as correct answer.

Satvik Golechha - 6 years, 5 months ago
Rajen Kapur
Jul 19, 2014

Let x = at, y = bt and z = ct (given) where t^3 = xyz/abc.then the given expression is sqrt(t^3).

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Shouldnt it be b^2y^2?? @Rishabh

Krishna Ar - 6 years, 10 months ago

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b^2y^2 isgiven

Rishabh Jain - 6 years, 10 months ago

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Yes @Rishabh Jain -but earlier it was b^2x^2. Are you preparing for JEE or Mathsolympiads?

Krishna Ar - 6 years, 10 months ago

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