Answer:
180
Explanation:
Consider the equation x + y + z = 15
From algebraic identities, we know that (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)
So,
(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + xz)
From the question, x^2 + y^2 + z^2 = 83 and x + y + z = 15
So,
152 = 83 + 2(xy + yz + xz)
=> 225 – 83 = 2(xy + yz + xz)
Or, xy + yz + xz = 142/2 = 71
Using algebraic identity a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca),
x3 + y3 + z3 – 3xyz = (x + y + z)(x² + y² + z² – (xy + yz + xz))
Now,
x + y + z = 15, x² + y² + z² = 83 and xy + yz + xz = 71
So, x^3 + y^3 + z^3 – 3xyz = 15(83 – 71)
=> x^3 + y^3 + z^3 – 3xyz = 15 × 12
Or, x^3 + y^3 + z^3 – 3xyz = 180
Answered by GAUTHMATH