Question

Prove that the value of $a^3+b^3+c^3-3abc$ is unaltered if we substitute $s-a,s-b,s-c$ for $a,b,c$ respectively, where $3s=2(a+b+c)$.

Answer 1

Given that:

$3s=2(a+b+c)$

Now,

$a^3+b^3+c^3-3abc=\frac{1}{2}(a+b+c)\left\{\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2\}$

When we substitute, $s-a,s-b,s-c$ for $a,b,c$ we get,

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

or, $2(a+b+c)-(a+b+c)=a+b+c$

$(s-a-s+b)=b-a$

Similarly,

$(s-b-s+c)=c-b$ and $(s-c-s+a)=a-c$

Substituting $s-a,s-b,s-c$ for $a,b,c$ we get,

$=\frac{1}{2}(s-a+s-b+s-c)\left\{\right(s-a-s+b{)}^2+(s-b-s+c{)}^2+(s-c-s+a{)}^2\}$

$=\frac{1}{2}(a+b+c)\left\{\right(b-a{)}^2+(c-b{)}^2+(a-c{)}^2\}$

$=\frac{1}{2}(a+b+c)\left\{\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2\}$

Therefore, Value of the expression will not altered.