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

Dear Student

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),

x^3+ y^3+ z^3– 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

x^3+ y^3+ z^3– 3xyz = 180

Thanks