Question

17. If x + y +Z=XYZ , prove that
(ii) x (1 - y2) (1 – 22) + y (1 - z2) (1 -- x2) + 2 (1 -- x2)(1 -- y2) = 4xyz.​

Answer 1

GIVEN

$⇒ x + y + z = xyz$

FORMULA

$x(y²+z²) + y(z²+x²)+z(x²+y²)+3xyz$

$=(x+y+z)(xy+yz+zx)$

PROOF

LHS

$= x (1 - y²) (1 – z²)$

$+ y (1 - z²) (1 - x²)$

$+ z(1 -x²)(1 - y²)$

$= x( 1 - y² - z² + y²z²)$

$+ y( 1 - z² - x² + z²x²)$

$+ z( 1 - x² - y² + x²y²)$

$= (x+y+z) - [ x(y²+z²)$

$+ y(z²+x²)+z(x²+y²)]$

$+ (xy²z² + yz²x² + zx²y²)$

$= (x+y+y) - [ x(y²+z²) + y(z²+x²)</p><p>+z(x²+y²) + 3xyz] + 3xyz </p><p>+ xyz(yz+zx+xy)$

$= (x+y+z) - (x+y+z)(xy+yz+zx)$ $+ 3xyz + xyz(xy+yz+zx)$

$= xyz - xyz(xy+yz+zx)$

$+ 3xyz + xyz(xy+yz+zx)$

$= xyz + 3xyz$

$= 4xyz$

= RHS

Hence Proved