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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Updated on: 10-Oct-2022

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