Question

(m-n) ³+(n-r) ³+(r-m) ³/6(m-n) (n-r) (r-m) =

Answer 1

Answer:

$\bf\frac{(m-n)^3+(n-r)^3+(r-m)^3}{6(m-n)(n-r)(r-m)}=\frac{1}{2}$

Step-by-step explanation:

$\textbf{Given:}$

$\displaystyle\frac{(m-n)^3+(n-r)^3+(r-m)^3}{6(m-n)(n-r)(r-m)}$

$\text{we know that,}$

$\text{If \bf\,a+b+c=0 , then \bf\,a^3+b^3+c^3=3abc }$

$\text{Here, (m-n)+(n-r)+(r-m)=0 }$

$\implies(m-n)^3+(n-r)^3+(r-m)^3=3(m-n)(n-r)(r-m)$

$\text{Now,}$

$\displaystyle\frac{(m-n)^3+(n-r)^3+(r-m)^3}{6(m-n)(n-r)(r-m)}$

$=\displaystyle\frac{3(m-n)(n-r)(r-m)}{6(m-n)(n-r)(r-m)}$

$=\frac{3}{6}$

$=\frac{1}{2}$

$\implies\boxed{\bf\frac{(m-n)^3+(n-r)^3+(r-m)^3}{6(m-n)(n-r)(r-m)}=\frac{1}{2}}$

Answer 2

So, The answer is $1$

Step-by-step explanation:

Given,

$\frac{(m-n)^{3}+(n-r)^{3}+(r-m)^{3} }{6(m-n)(n-r)(r-n)} =?$

We know,

$a^{3}+b^{3}+c^{3}-3abc=0$ if $(a+b+c)=0$

∴$a^{3}+b^{3}+c^{3}=3abc$

Here,

$(m-n)+(n-r)+(r-m)=0$

Then,

$(m-n)^3+(n-r)^3+(r-m)^3=3(m-n)(n-r)(r-m)$

⇒$\frac{(m-n)^{3}+(n-r)^{3}+(r-m)^{3} }{3(m-n)(n-r)(r-n)} =1$

∴$\frac{(m-n)^{3}+(n-r)^{3}+(r-m)^{3} }{6(m-n)(n-r)(r-n)} =\frac{1}{2}$  (by multiplying $\frac{1}{2}$ on both sides )

∴ The answer is $1$