Question
Factorise:
$(a-3 b)^3+(3 b-c)^3+(c-a)^3$
$(a-3 b)^3+(3 b-c)^3+(c-a)^3$
✓
Answer
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, $x=(a-3 b), y=(3 b-c), z=(c-a)$
$(a-3 b)^3+(3 b-c)^3+(c-a)^3$
$=(a-3 b+3 b-c+c-a)$
${\left[(a-3 b)^2+(3 b-c)^2+(c-a)^2-(a-3 b)(3 b-c)-(3 b-c)(c-a)-(c-a)(a-3 b)\right]}$
$+3(a-3 b)(3 b-c)(c-a)$
$=0+3(a-3 b)(3 b-c)(c-a)$
$=3(a-3 b)(3 b-c)(c-a)$
$x^3+y^3+z^3-3 x y z=(x+y+z)\left(x^2+y^2+z^2-x y-y z-z x\right)$
$x^3+y^3+z^3=(x+y+z)\left(x^2+y^2+z^2-x y-y z-z x\right)+3 x y z$
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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