Question
Find the following products: $(2a - 3b - 2c)(4a^2 + 9b^2 + 4c^2 + 6ab - 6bc + 4ca)$
✓
Answer
We have,
$(2a - 3b - 2c)(4a^2 + 9b^2 + 4c^2 + 6ab - 6bc + 4ca)$
$= (2a + (-3b) + (-2c)) + \big((2a)^2 + (-3b)^2 + (-2c)^2 - (2a)(-3b)(-2c) - (-2c)(2a)\big)$
$= (2a)^3 + (-3b)^3 + (-2c)^3 - 3(2a)(-3b)(-2c)$
$\big[\because a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\big]$
$= 8a^3 - 27b^3 - 8c^3 - 36abc$
$\therefore (2a - 3b - 2c)(4a^2 + 9b^2 + 4c^2 + 6ab - 6bc + 4ca)$
$ = 8a^3 - 27b^3 - 8c^3 - 36abc$
$(2a - 3b - 2c)(4a^2 + 9b^2 + 4c^2 + 6ab - 6bc + 4ca)$
$= (2a + (-3b) + (-2c)) + \big((2a)^2 + (-3b)^2 + (-2c)^2 - (2a)(-3b)(-2c) - (-2c)(2a)\big)$
$= (2a)^3 + (-3b)^3 + (-2c)^3 - 3(2a)(-3b)(-2c)$
$\big[\because a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\big]$
$= 8a^3 - 27b^3 - 8c^3 - 36abc$
$\therefore (2a - 3b - 2c)(4a^2 + 9b^2 + 4c^2 + 6ab - 6bc + 4ca)$
$ = 8a^3 - 27b^3 - 8c^3 - 36abc$
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