Example 32

Transcript

Example 32 Show that = + 2 + 2 + 2 = 2xyz (x + y + z)3 Taking L.H.S = + 3 + 2 + 2 Divide & Multiply by xyz = + 3 + 2 + 2 = 1 x. y. z + 3 + 2 + 2 = 1 x. y. z + 3 + 2 + 2 Multiplying R1 by x , R2 by y & R3 by z = 1 + 2 ( ) ( ) ( ) + 2 ( ) ( ) 2 + 2 Taking out x common from C1, y common from C2 & z common from C3 = y+z 2 x2 x2 y2 x+z 2 y2 2 z2 x+y 2 Applying C2 C2 C1 = y+z 2 x2 + 2 x2 y2 x+z 2 y2 y2 2 z2 z2 x+y 2 = y+z 2 (x + )( + + ) x2 y2 (x+z) ( + + ) y2 2 0 x+y 2 = y+z 2 ( )( + + ) x2 y2 x+z y ( + + ) y2 2 0 x+y 2 Taking out ( + + ) common from C2 = ( + + ) y+z 2 x2 y2 + y2 2 0 x+y 2 Applying C3 C3 C1 = ( + + ) y+z 2 2 + 2 y2 + 2 2 2 0 + 2 2 = ( + + ) y+z 2 ( + + )( ( + )) y2 + 0 2 0 ( + + )(( + ) ) Taking out ( + + ) Common from C3 = ( + + )( + + ) y+z 2 y2 + 0 2 0 + = + + 2 y+z 2 y2 + 0 2 0 + Applying R1 R1 R2 R3 = + + 2 y+z 2 2 2 + 0 0 ( + ) y2 + 0 2 0 + = + + 2 2+ 2+2 2 2 + + y2 + 0 2 0 + = + + 2 2 2 2 y2 + 0 2 0 + Applying C2 C2 + 1 C1 = + + 2 2 2 + ( ) 2 y2 x + + ( ) 0 2 0+ ( ) + = + + 2 2 0 2 y2 x+ 0 2 2 + Applying C3 C3 + 1 C1 = + + 2 2 0 2 + ( ) y2 + 0+ ( ) 2 2 + + ( ) = + + 2 2 0 0 y2 + 2 2 2 + Expanding Determinant along R1 = + + 2 2 + 2 2 + 0 2 2 2 + +0 2 + 2 2 = + + 2 2 + 2 2 + 0+0 = + + 2 2yz (x + z) (x + y) 2 2 0 + 0 = + + 2 (2yz ((x + z) (x + y) zy ) = + + 2 (2yz) ((x + z) (x + y) zy ) = + + 2 (2yz) (x2 + xy + zx + zy zy) = + + 2 (2yz) (x2 + xy + xz) = + + 2 (2yz) . x (x + y + z) = + + 2 (2xyz) = (2xyz) + + 2 = R.H.S Hence Proved