Question

If $a+b+c=8$, then the value of ,${(a-4)}^3+{(b-3)}^3+{(c-1)}^3-3(a-4)(b-3)(c-1)$ is:

• A. $2$
• B. $4$
• C. $1$
• D. $0$

Answer 1

The correct option is D $0$

Given, $a+b+c=8$

$\Rightarrow a+b+c-3=0$

$\Rightarrow (a-4)+(b-3)+(c-1)=\mathrm{0...}\left(1\right)$

We know, $a^3+b^3+c^3=3abc$

if $a+b+c=0$ or $a=b=c$

So, from (1) we can say

$(a-4{)}^3+(b-3{)}^3+(c-1{)}^3=3(a-4\left)\right(b-3\left)\right(c-1)$

$\Rightarrow (a-4{)}^3+(b-3{)}^3+(c-1{)}^3-3(a-4\left)\right(b-3\left)\right(c+1)=0$

Hence, ans $=0$