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Statement

If x4+y4+z4=0...

Question

If $x^{4} + y^{4} + z^{4} = 0,$ then $∣ ∣ ∣ ∣ \begin{matrix} 1 & x y & y z z x & 1 & x y y z & z x & 1 \end{matrix} ∣ ∣ ∣ ∣ =$ (where $x, y, z \in R)$

A

0

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B

x + y + z + 3

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C

1

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D

x y x + 2

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Solution

The correct option is C $1$
Since $x, y, z \in R$
and $x^{4} + y^{4} + z^{4} = 0$
$\Rightarrow x = y = z = 0$

$\Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣ = 1$

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Similar questions

x^{4} + y^{4} + z^{4} = 0

∣ ∣ ∣ ∣ \begin{matrix} 1 & x y & y z z x & 1 & x y y z & z x & 1 \end{matrix} ∣ ∣ ∣ ∣ =

___________ (where x, y, z

\in

Q. The number of triples

(x, y, z)

of real numbers satisfying the equation

x^{4} + y^{4} + z^{4} + 1 = 4 x y z

{2 x}^{4} = y^{4} + z^{4}

x y z = 8

{l o g}_{y} x, {l o g}_{z} y, {l o g}_{x} z

G . P .

, then find the value of

x, y, z

R

s i n^{- 1} x + s i n^{- 1} y + s i n^{- 1} z = π

x^{4} + y^{4} + z^{4} + 4 x^{2} y^{2} z^{2} = k (x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2})

k

If $s i n^{- 1} x + s i n^{- 1} y + s i n^{- 1} z = π$ then $x^{4} + y^{4} + z^{4} + 4 x^{2} y^{2} z^{2} = k (x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2})$ where k is equal to

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