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Find the following products:$(3x + 2y + 2z) (9x^2 + 4y^2 + 4z^2 – 6xy – 4yz – 6zx)$
Given:
$(3x + 2y + 2z) (9x^2 + 4y^2 + 4z^2 – 6xy – 4yz – 6zx)$
To do:
We have to find the given product.
Solution:
We know that
$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=a^3+b^3+c^3-3abc$
Therefore,
$(3x + 2y + 2z) (9x^2 + 4y^2 + 4z^2 – 6xy – 4yz – 6zx)= (3x + 2y + 2z) [(3x)^2 + (2y)^2 + (2z)^2 – 3x \times 2y + 2y \times 2z + 2z \times 3x]$
$= (3x)^3 + (2y)^3 + (2z)^3 – 3 \times 3x \times 2y \times 2z$
$= 27x^3 + 8y^3 + 8z^3 – 36xyz$
Hence, $(3x + 2y + 2z) (9x^2 + 4y^2 + 4z^2 – 6xy – 4yz – 6zx)=27x^3 + 8y^3 + 8z^3 – 36xyz$.
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