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Question

Show that the product of $${ a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }-3abc$$ and $${ x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 }-3xyz$$ can be put into the form $${ A }^{ 3 }+{ B }^{ 3 }+{ C }^{ 3 }-3ABC$$.


Solution

Let the product of $$a^3+b^3+c^3-3abc$$ and $$x^3+y^3+z^3-3xyz$$ as $$(a+b+c)(a+wb+w^2c)(a+w^2b+wc)\times(x+y+z)(x+wy+w^2z)(x+w^2y+wz)$$.
By taking six factors in the pair;
$$(a+b+c)(x+y+z); (a+wb+w^2c)(x+wy+w^2z); (x+w^2y+wz)(a+w^2b+wc) $$
BY using partial products concept, we get
$$(a+b+c)(a+wb+w^2c)(a+w^2b+wc)\times(x+y+z)(x+wy+w^2z)(x+w^2y+wz)$$
$$=(A+B+C)(A+wB+w^2C)(A+w^2B+WC)$$
By comparing equal terms of $$w, w^2$$ on both sides, we get the values of $$A, B$$ and $$C$$ as
$$A=ax+by+cz, B=bx+cy+az, C=cx+ay+bz$$
Thus, we can write the product of $$(A+B+C)(A+wB+w^2C)(A+w^2B+wC)=A^3+B^3+C^3-3ABC$$


Mathematics

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