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