Find the following products:$(2a – 3b – 2c) (4a^2 + 9b^2 + 4c^2 + 6ab – 6bc + 4ca)$

Given: 

$(2a – 3b – 2c) (4a^2 + 9b^2 + 4c^2 + 6ab – 6bc + 4ca)$

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,

$(2a -3b- 2c) (4a^2 + 9b^2 + 4c^2 + 6ab – 6bc + 4ca)= (2a -3b- 2c) [(2a)^2 + (-3b)^2 + (-2c)^2 – 2a \times (-3b) – (-3b) \times (-2c) – (-2c) \times 2a]$

$= (2a)^3 + (-3b)^3 + (-2c)^3 -3 \times 2a \times (-3 b) \times (-2c)$

$= 8a^3 – 27b^3 -8c^3 – 36abc$

Hence, $(2a -3b- 2c) (4a^2 + 9b^2 + 4c^2 + 6ab – 6bc + 4ca)=8a^3 – 27b^3 -8c^3 – 36abc$.

Tutorialspoint

Updated on: 10-Oct-2022

44 Views

Kickstart Your Career

Get certified by completing the course

Get Started