If x+y+z = 1, x^2 +y^2 +z^2 = 3, x^3 +y^3 +z^3 = 7, Then, x^5 +y^5 +z^5 = ?

Open in App

Solution

Suggest Corrections

Similar questions

x, y, z

x + y + z = 0, x^{2} + y^{2} + z^{2} = a^{2}, x^{3} + y^{3} + z^{3} = b^{3}, x^{5} + y^{5} + z^{5} = c^{5}

x, y, z \in + R

x^{2} + y^{2} + z^{2} = 27

x^{3} + y^{3} + z^{3}

x^{3} + y^{3} + z^{3} - 3 x y z = \frac{1}{2} (x + y + z) [(x - y)^{2} + (y - z)^{2} + (z - x)^{2}]

x^{3} + y^{3} + z^{3} - 3 x y z = \frac{1}{2} (x + y + z) [(x - y)^{2} + (y - z)^{2} + (z - x)^{2}]

Verify that : $x^{3} + y^{3} + z^{3} - 3 x y z = \frac{1}{2} (x + y + z) [{(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2})]$

Join BYJU'S Learning Program