Factorize a3b-c3+b3c-a3+c3a-b3.

# Factorize ${a}^{3}{\left(b-c\right)}^{3}+{b}^{3}{\left(c-a\right)}^{3}+{c}^{3}{\left(a-b\right)}^{3}$.

1. A
$3\mathit{abc}\left(b-c\right)\left(c+a\right)\left(a-b\right)$
2. B
$3\mathit{abc}\left(b-c\right)\left(c-a\right)\left(a-b\right)$
3. C
$3\mathit{abc}\left(b-c\right)\left(c-a\right)\left(a+b\right)$
4. D
$3\mathit{abc}\left(b-c\right)\left(c-a\right)\left(b-a\right)$

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### Solution:

Given, ${a}^{3}{\left(b-c\right)}^{3}+{b}^{3}{\left(c-a\right)}^{3}+{c}^{3}{\left(a-b\right)}^{3}$.
We know the formula,
${x}^{3}+{y}^{3}+{z}^{3}-3\mathit{xyz}=\left(x+y+z\right)\left({x}^{2}+{y}^{2}+{z}^{2}-\mathit{xy}-\mathit{yz}-\mathit{zx}\right)$ .
If then ${x}^{3}+{y}^{3}+{z}^{3}=3\mathit{xyz}$.
Therefore, rewrite the given expression we have,
${a}^{3}{\left(b-c\right)}^{3}+{b}^{3}{\left(c-a\right)}^{3}+{c}^{3}{\left(a-b\right)}^{3}$
$={\left[a\left(b-c\right)\right]}^{3}+{\left[b\left(c-a\right)\right]}^{3}+{\left[c\left(a-b\right)\right]}^{3}$
$={\left(\mathit{ab}-\mathit{ac}\right)}^{3}+{\left(\mathit{bc}-\mathit{ab}\right)}^{3}+{\left(\mathit{ac}-\mathit{bc}\right)}^{3}$
Here, we have $x=\mathit{ab}-\mathit{ac}$, $y=\mathit{bc}-\mathit{ab}$, and $z=\mathit{ac}-\mathit{bc}$.
Therefore, adding x, y and z, we get,
$x+y+z=\left(\mathit{ab}-\mathit{ac}\right)+\left(\mathit{bc}-\mathit{ab}\right)+\left(\mathit{ac}-\mathit{bc}\right)$
$⇒\mathit{ab}-\mathit{ac}+\mathit{bc}-\mathit{ab}+\mathit{ac}-\mathit{bc}=0$
${a}^{3}{\left(b-c\right)}^{3}+{b}^{3}{\left(c-a\right)}^{3}+{c}^{3}{\left(a-b\right)}^{3}$
$⇒{\left(\mathit{ab}-\mathit{ac}\right)}^{3}+{\left(\mathit{bc}-\mathit{ab}\right)}^{3}+{\left(\mathit{ac}-\mathit{bc}\right)}^{3}$
$⇒3\left(\mathit{ab}-\mathit{ac}\right)\left(\mathit{bc}-\mathit{ab}\right)\left(\mathit{ac}-\mathit{bc}\right)$
$⇒3\left[a\left(b-c\right)\right]\left[b\left(c-a\right)\right]\left[c\left(a-b\right)\right]$
$⇒3\mathit{abc}\left(b-c\right)\left(c-a\right)\left(a-b\right)$
Hence, the required answer $3\mathit{abc}\left(b-c\right)\left(c-a\right)\left(a-b\right)$.
Therefore, option 2 is correct.

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