Question

If x+ y+z=15 andx 2 + y 2 +z 2=33, find the value of x 3 + y 3 + z 3−3 xyz

Answer 1

Answer:

180

Step-by-step explanation:

(x+y+z)2= x2+y2+z2+2(xy+yz+zx)

15 2= x2+y2+z2+xy+yz+zx

225= x2+y2+z2+xy+yz+zx

225= (33) + xy+yz+zx

2(xy+yz+zx)= 225-83=142

xy+yz+zx= 71

x3+y3+z3-3xyz= (x+y+z)(x2+y2+z2-xy-yz-zx)  (formula)

                        = (15)(83+ -1(xy+yz+zx))

                        =(15)(83-71)

                       = 15*12

                        =180

Also, I HAVE SEEN THIS QUESTION BEFORE AND I THINK THE NUMBER IS 83 NOT 33 IF IT WERE 33 THE ANSWER WOULD GO INTO NEGATIVE. PLEASE MARK BRAINLIEST.