If x + y + z = 3, $x^2 + y^2 + z^2 = 45$ and $x^3 + y^3 + z^3 = 69,$ then what is the value of xyz ? |
-40 40 -31 30 |
-40 |
x + y + z = 3, $x^2 + y^2 + z^2 = 45$ $x^3 + y^3 + z^3 = 69,$ then what is the value of xyz= ? We know that, (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx) x3 + y3 + z3 – 3xyz = (x + y + z){( x2 + y2 + z2 ) – (xy + yz +zx)} Now, = (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx) = (3)2 = 45 + 2(xy + yz +zx) = 9 = 45 + 2(xy + yz +zx) = -36 = 2(xy + yz +zx) = (xy + yz +zx) = -18 Now, x3 + y3 + z3 – 3xyz = (x + y + z){( x2 + y2 + z2 ) – (xy + yz +zx)} = 69 – 3xyz = 3 {(45) – (-18)} = 69 – 3xyz = 3 × {(45 + 18)} = 69 – 3xyz = 3 × 63 = 69 – 3xyz = 189 = 3xyz = (69 – 189) = 3xyz = -120 = xyz = -40 |