Question

Find the value of $(x-a{)}^3+(x-b{)}^3+(x-c{)}^3-3(x-a\left)\right(x-b\left)\right(x-c)$ when $a+b+c=3x$.

Answer 1

$(x-a{)}^3+(x-b{)}^3+(x-c{)}^3-3(x-a\left)\right(x-b\left)\right(x-c)$ and $a+b+c=3x$
We know that $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$
If $(a+b+c)=0$, then.$a^3+b^3+c^3=3abc$
Here,$a=x-a$
$b=x-b$
$c=x-c$ and $a+b+c=x-a+x-b+x-c=3x-(a+b+c)=3x-3x=0$
Thus,$(x-a{)}^3+(x-b{)}^3+(x-c{)}^3-3(x-a\left)\right(x-b\left)\right(x-c)$
$=0$