Harshit Singh
Last Activity: 4 Years ago
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