If $x + y + z = 4, x^{2} + y^{2} + z^{2} = 14$ and $x^{3} + y^{3} + z^{3} = 34$ , then the number of ordered triplet solution (x, y, z) is given by

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Solution

The correct option is C
6

Consider a polynomial $P (t) = t^{3} + a t^{2} + b t + c$ having roots x, y, z

Clearly a = - 4, b = 1 has roots x, y, z

$∴ t^{3} - 4 t^{2} + t + c = 0$ has roots x, y, z

$\Rightarrow (x^{3} + y^{3} + z^{3}) - 4 (x^{2} + y^{2} + z^{2}) + (x + y + z) + 3 c = 0$

$\Rightarrow c = 6$

$∴ P (t) = (t + 1) (t - 2) (t - 3)$

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Similar questions

x + y + z = 4; x^{2} + y^{2} + z^{2} = 14; x^{3} + y^{3} + z^{3} = 34

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

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∣ ∣ ∣ ∣ ∣ \begin{matrix} f_{x} & f_{y} & f_{z} g_{x} & g_{y} & g_{z} h_{x} & h_{y} & h_{z} \end{matrix} ∣ ∣ ∣ ∣ ∣ = ?

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}]

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