If `x+y+z=xyz`, show that : `(3x-x^3)/(1-3x^2) + (3y-y^3)/(1-3y^2) + (3z-z^3)/(1-3z^2) = (3x-x^3)/(1-3x^2) . (3y-y^3)/(1-3y^2) . (3z-z^3)/(1-3z^2)`

If x + y + z = xyz , prove that (3x -x^(3))/ (1-3x^(2)) + (3y -y^(3))/(1- 3y^(2)) +(3z -z^(3))/(1- 3z^(2)) = (3x -x^(3))/(1-3x)^(2) * (3y- y^(3))/(1-3x)^(2)* (3z- z^(3))/(1-3z)^(2) .

(3x-x^(3))/(1-3x^(2))+(3y-y^(3))/(1-3y^(2))+(3z-z^(3))/ (1-3z^(2))=((3x-x^(3))/(1-3x^(2)))((3z-z^(3))/(1-3z^(2) ))(((3y-y^(3))/(1-3y^(2))))

Find ((x^(2)-y^(2))^(3) + (y^(2) -z^(2))^(3)+ (z^(2) -x^(2))^(3))/((x-y)^(3) + (y-z)^(3) + (z-x)^(3))

If x+y+z=6, then the value of (x-1)^(3)+(y-2)^(3)+(z-3)^(3) is

Find the S.D. between the lines : (i) (x)/(2) = (y)/(-3) = (z)/(1) and (x -2)/(3) = (y - 1)/(-5) = (z + 4)/(2) (ii) (x -1)/(2) = (y - 2)/(3) = (z - 3)/(2) and (x + 1)/(3) = (y - 1)/(2) = (z - 1)/(5) (iii) (x + 1)/(7) = (y + 1)/(-6) = (z + 1)/(1) and (x -3)/(1) = (y -5)/(-2) = (z - 7)/(1) (iv) (x - 3)/(3) = (y - 8)/(-1) = (z-3)/(1) and (x + 3)/(-3) = (y +7)/(2) = (z -6)/(4) .

The image of the line (x-1)/(3)=(y-3)/(1)=(z-4)/(-5) in the plane 2x-y+z+3=0 is the line (1)(x+3)/(3)=(y-5)/(1)=(z-2)/(-5) (2) (x+3)/(-3)=(y-5)/(-1)=(z+2)/(5) (3) (x-3)/(3)=(y+5)/(1)=(z-2)/(-5) (3) (x-3)/(-3)=(y+5)/(-1)=(z-2)/(5)