2 equations and 3 variables?
y x + z y + x z = 7 x y + y z + z x = 9
If x , y , z are non zero real numbers such that they satisfy the equations above, find the value of: y 3 x 3 + z 3 y 3 + x 3 z 3 − 4
The answer is 153.
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4 solutions
Same approach!!
Take the three terms of first equation as a,b,c. Now use the identity:
a^3+b^3+c^3-3abc=[(a+b+c)^2-3(ab+bc+ca)]
This question is from KVPY , isn't it ?
H e r e ′ s a n o t h e r m e t h o d t o s o l v e i t :
G i v e n , y x + z y + x z = 7 . . . . . . . . . . . . . . . . . . . . . ( 1 ) x y + y z + z x = 9 . . . . . . . . . . . . . . . . . . . . . ( 2 )
W e k n o w t h a t , ( y x + z y + x z ) 3 = ( y x ) 3 + ( z y ) 3 + ( x z ) 3 + 3 ( x 2 y z ) + 3 ( y z x 2 ) + 3 ( y 2 x z ) + 3 ( x z y 2 ) + 3 ( z 2 x y ) + 3 ( x y z 2 ) + 6 . . . . . . . . . . . ( 3 )
On\quad multiplying\quad (1)\quad and\quad (2)\quad we\quad get,\\ \left( \frac { x }{ y } +\frac { y }{ z } +\frac { z }{ x } \right) \left( \frac { y }{ x } +\frac { z }{ y } +\frac { x }{ z } \right) =3+\frac { yz }{ x^{ 2 } } +{ \frac { x^{ 2 } }{ yz } }+\frac { xz }{ y^{ 2 } } +\frac { y^{ 2 } }{ xz } +\frac { xy }{ z^{ 2 } } +\frac { z^{ 2 } }{ xy } =63\\ 60=\frac { yz }{ x^{ 2 } } +{ \frac { x^{ 2 } }{ yz } }+\frac { xz }{ y^{ 2 } } +\frac { y^{ 2 } }{ xz } +\frac { xy }{ z^{ 2 } } +\frac { z^{ 2 } }{ xy } \\ \\ On\quad replacing\quad this\quad value\quad in\quad (3),\\ (7)^{ 3 }={ \left( \frac { x }{ y } \right) }^{ 3 }+{ \left( \frac { y }{ z } \right) }^{ 3 }+{ \left( \frac { z }{ x } \right) }^{ 3 }+180+6\\ \\ { \left( \frac { x }{ y } \right) }^{ 3 }+{ \left( \frac { y }{ z } \right) }^{ 3 }+{ \left( \frac { z }{ x } \right) }^{ 3 }=157\\ \\ \therefore \quad { \left( \frac { x }{ y } \right) }^{ 3 }+{ \left( \frac { y }{ z } \right) }^{ 3 }+{ \left( \frac { z }{ x } \right) }^{ 3 }-4=\boxed{153}
Yes, I feel like I had practiced it before from a question paper of KVPY or any such exam.
( y x + z y + x z ) 2 ⇒ y 2 x 2 + z 2 y 2 + x 2 z 2 = y 2 x 2 + z 2 y 2 + x 2 z 2 + 2 ( y x ˙ z y + z y ˙ x z + x z ˙ y x ) = y 2 x 2 + z 2 y 2 + x 2 z 2 + 2 ( z x + x y + y z ) = ( y x + z y + x z ) 2 − 2 ( z x + x y + y z ) = 7 2 − 2 × 9 = 4 9 − 1 8 = 3 1
Now, we have:
\(\begin{array} {} \dfrac {x^3}{y^3} + \dfrac {y^3}{z^3} + \dfrac {z^3}{x^3} -4 \\ = \left( \dfrac {x}{y} + \dfrac {y}{z} + \dfrac {z}{x} \right) \left( \dfrac {x^2}{y^2} + \dfrac {y^2}{z^2} + \dfrac {z^2}{x^2} \right) - \left( \dfrac {x}{z} + \dfrac {y}{x} + \dfrac {z}{y} \right) \left( \dfrac {x}{y} + \dfrac {y}{z} + \dfrac {z}{x} \right) \\ \quad + 3 \left( \dfrac {x}{y} \dot{} \dfrac {y}{z} \dot{} \dfrac {z}{x} \right) - 4 \\ = 7\times 31 - 9\times 7 + 3 - 4 = \boxed{153} \end{array}\)