Find the value of x³+ y³ + z³ – 3xyz if x² + y² + z² = 83 and x + y + z = 15​

identity used is

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

then use

(x+y+z)² = x²+y²+z²+2(xy+yz+xz)

225= 83 + 2(xy+yz+xz)

xy+yz+xz = (225-83)/2

xy+yz+xz= 142/2

xy+yz+xz= 71

ok

now use identity

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

now

x³ + y³+ z³– 3xyz = 15 (83 – xy – yz – zx).

= 15[83 - (71)]

= 15×12

=180

see explanation

step-by-step explanation:

a

(6r + 7) + (13 + 7r) ( remove brackets and collect like terms )

= 6r + 7 + 13 + 7r

= 13r + 20

b

remove brackets and collect like terms

13 - r - 1 + r

= 12 - r

c

remove brackets and collect like terms

= - 8 - r + 2r - 4

= - 12 + r

d

remove brackets and collect like terms

= 7r - - - 6r

[-   - = - - = - ]

= r -