Shew that $(a^{2} + b^{2} + c^{2} - b c - c a - a b) (x^{2} + y^{2} + z^{2} - z^{2} - y z - z x - x y)$ may be put into the form $A^{2} + B^{2} + C^{2} - B C - C A - A B$ .

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Solution

It is known that $a^{2} + b^{2} + c^{2} - a b - b c - c a = (a + w b + w^{2} c) (a + w c + w^{2} b)$
$∴ (a^{2} + b^{2} + c^{2} - a b - b c - c a) (x^{2} + y^{2} + z^{2} - x y - y z - z x) = (a + w b + w^{2} c) (a + w c + w^{2} b) \times (x + y w + w^{2} z) (x + w z + w^{2} y)$
By taking these $4$ quantities in pairs, we have $(a + w b + w^{2} c) (x + y w + w^{2} z)$ and $(a + w c + w^{2} b) (x + w z + w^{2} y)$
we obtain true partial products
$(A + w B + w^{2} C) (A + w^{2} B + w C) = A^{2} + B^{2} + C^{2} - A B - B C - C A$

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