Simplify: $a^{3} (b - c)^{3} + b^{3} (c - a)^{3} + c^{3} (a - b)^{3}$

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Solution

$= a^{3} (b - c)^{3} + b^{3} (c - a)^{3} + c^{3} (a - b)^{3}$
$= (a b - a c)^{3} + (b c - b a)^{3} + (c a - c b)^{3}$
We know that ,
$x^{3} + y^{3} + z^{3} = 3 x y z$ $i f$ $x + y + z = 0$
So , here
$(a b - a c) + (b c - b a) + (c a - c b) = 0$
Hence ,
$= a^{3} (b - c)^{3} + b^{3} (c - a)^{3} + c^{3} (a - b)^{3}$
$= 3 (a b - a c) (b c - b a) (c a - c b)$
$= 3 a b c (b - c) (c - a) (a - b)$

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Similar questions

a^{3} {(b - c)}^{3} + b^{3} {(c - a)}^{3} + c^{3} {(a - b)}^{3}

{(a + b + c)}^{3} - a^{3} - b^{3} - c^{3} = 3 (a + b) (b + c) (c + a)

{(a + b + c)}^{3} - a^{3} - b^{3} - c^{3} = [{(a + b + c)}^{3} - a^{3}] - [b^{3} + c^{3}]

a^{3} {(b - c)}^{3} + b^{3} {(c - a)}^{3} + c^{3} {(a - b)}^{3} =

a^{3} (c - b) + b^{3} (a - c) + c^{3} (b - a)

> 0

\frac{a^{3}}{b^{3}} + \frac{b^{3}}{c^{3}} + \frac{c^{3}}{a^{3}} \geq 3

∣ ∣ ∣ ∣ \begin{matrix} a - b & b + c & a b - c & c + a & b c - a & a + b & c \end{matrix} ∣ ∣ ∣ ∣

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