Find cyclic factors:
$x (y^{2} + z^{2}) + y (z^{2} + x^{2}) + z (x^{2} + y^{2}) + 2 x y z$

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Solution

Given: $x (y^{2} + z^{2}) + y (z^{2} + x^{2}) + z (x^{2} + y^{2}) + 2 x y z$
Expanding the equation we have,
$x y^{2} + x z^{2} + y z^{2} + y x^{2} + z x^{2} + z y^{2} + 2 x y z$
Rearranging the above equation
$x y^{2} + z y^{2} + x y z + y x^{2} + z x^{2} + x y z + x z^{2} + y z^{2}$
$(x y^{2} + z y^{2} + x y z) + (y x^{2} + z x^{2} + x y z) + (x z^{2} + y z^{2})$
Taking commons,
$y (x y + z y + x z) + x (x y + x z + y z) + z^{2} (x + y)$
Taking $(x y + z y + x z)$ common from first two terms
$(y + x) (x y + z y + x z) + z^{2} (x + y)$
$(x + y) (x y + z y + x z) + z^{2} (x + y)$
Taking $(x + y)$ common
$(x + y) (x y + z y + x z + z^{2})$
$(x + y) ((x y + z y) + (x z + z^{2}))$
Taking $y$ and $z$ common
$(x + y) (y (x + z) + z (x + z))$
Taking $(x + z)$ common
$(x + y) (y + z) (x + z)$
Hence the cyclic factors of the given equation are
$(x + y)$ , $(y + z)$ and $(x + z)$

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Write the following by using ∑ notation

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