Question

if x+y+z=9 & x²+y²+z²=35, find x³+y³+z³-3xyz​

Answer 1

Answer:

The answer is 108..

x+y+z = 9

x²+y²+z²=35

(x+y+z)² = x²+y²+z²+2xy+2zx+2yz

--> 9² = 35+2(xy+yz+zx)

--> 81-35 = 2(xy+yz+zx)

--> 46/2 = (xy+yz+zx)

--> -23 = -xy-yz-zx

So, with the equation,

x³+y³+z³-3xyz

Let's expand the equation and distribute its exponential powers,,

= (x+y+z)(x²+y²+z² -xy-yz-zx)

= 9*(35-23)

= 9*12

= 108

Hope it helped you..

Thank you !!