Find the following products:$(4x -3y + 2z) (16x^2 + 9y^2+ 4z^2 + 12xy + 6yz – 8zx)$

Given: 

$(4x -3y + 2z) (16x^2 + 9y^2+ 4z^2 + 12xy + 6yz – 8zx)$

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,

$(4x – 3y + 2z) (16x^2 + 9y^2 + 4z^2 + 12xy + 6yz – 8zx)= (4x -3y + 2z) [(4x)^2 + (-3y)^2 + (2z)^2 – 4x \times (-3y) - (-3y) \times (2z) – (2z) \times (4x)]$

$= (4x)^3 + (-3y)^3 + (2z)^3 – 3 \times 4x \times (-3y) \times (2z)$

$= 64x^3 – 27y^3 + 8z^3 + 72xyz$

Hence, $(4x – 3y + 2z) (16x^2 + 9y^2 + 4z^2 + 12xy + 6yz – 8zx)=64x^3 – 27y^3 + 8z^3 + 72xyz$.