Factorise:

Question:

Factorise:

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

 

Solution:

We know

$x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$

$x^{3}+y^{3}+z^{3}=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)+3 x y z$

Here, $x=(a-3 b), y=(3 b-c), z=(c-a)$

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

$=(a-3 b+3 b-c+c$

$-a)\left[(a-3 b)^{2}+(3 b-c)^{2}+(c-a)^{2}-(a-3 b)(3 b-c)-(3 b-c)(c-a)-(c-a)(a-3 b)\right]$

$+3(a-3 b)(3 b-c)(c-a)$

$=0+3(a-3 b)(3 b-c)(c-a)$

$=3(a-3 b)(3 b-c)(c-a)$

 

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