Show that ${(x + y + z)}^{2} - x^{5} - y^{5} - z^{5} = 5 (y + z) (z + x) (x + y) (x^{2} + y^{2} + z^{2} + y z + z x + x y)$ .

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Solution

Let $E = (x + y + z)^{2} - x^{5} - y^{5} - z^{5}$
If $y = - z$ , then $E = 0,$ that means $y + z$ is factor for $E$
Similarly $z + x, x + y$ are also factors of E
Also since $E$ is of the fifth degree the remaining factor is of the second degree, and, since it is symmetrical in $x, y, z$ , it must be of the form
$A (x^{2} + y^{2} + z^{2}) + B (y z + z x + x y)$ .
Put $x = y = z = 1$ ; thus $10 = A + B$ ....... $(i)$
put $x = 2, y = 1, z = 0$ ; thus $35 = 5 A + 2 B$ ....... $(i i)$
Solving $(i)$ and $(i i)$ , we get
$B = 5$ and $A = 5$
Therefore, $(x + y + z)^{2} - x^{5} - y^{5} - z^{5} = (x^{2} y + x^{2} z + y^{2} x + y z x + x y z + x z^{2} + y^{2} z + y z^{2}) (5 (x^{2} + y^{2} + z^{2}) + 5 (x y + y z + z x)$
Which on further simplification, we get
$(x + y + z)^{2} - x^{5} - y^{5} - z^{5} = 5 (x + y) (y + z) (x + z) ((x^{2} + y^{2} + z^{2}) + (x y + y z + z x))$

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