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Factorising a Binomial

Factorize the...

Question

Factorize the following:
$x^{2} y z + x y^{2} z + x y z^{2}$

x y z (x - y + z)

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x y (x + y + z)

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x z (x + y + z)

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x y z (x + y + z)

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Solution

The correct option is C $x y z (x + y + z)$
$x^{2} y z + x y^{2} z + x y z^{2}$
Common factor= $x y z$
$= x y z (x + y + z)$

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Similar questions

\frac{1}{x y z} (x^{2} + y^{2} + z^{2}) + 2 (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})

|\begin{matrix} z & x & y \\ z^{2} & x^{2} & y^{2} \\ z^{4} & x^{4} & y^{4} \end{matrix}| = |\begin{matrix} x & y & z \\ x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \end{matrix}| = |\begin{matrix} x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \\ x & y & z \end{matrix}| = x y z (x - y) (y - z) (z - x) (x + y + z) .

If y is the mean proportional between x and z prove that

xyz(x+y+z)^3=(xy+yz+zx)^3

y

x

z

x y z (x + y + z)^{3} = (x y + y z + z x)^{3}

A = [\begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix}]

[\begin{matrix} x^{- 1} & 0 & 0 \\ 0 & y^{- 1} & 0 \\ 0 & 0 & z^{- 1} \end{matrix}]

x y z [\begin{matrix} x^{- 1} & 0 & 0 \\ 0 & y^{- 1} & 0 \\ 0 & 0 & z^{- 1} \end{matrix}]

\frac{1}{x y z} [\begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix}]

\frac{1}{x y z} [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]

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