Question

Find the value of ${(\frac{1}{2}x-\frac{1}{3}y)}^3+{(\frac{1}{3}y-\frac{1}{4}z)}^3+{(\frac{1}{4}z-\frac{1}{2}x)}^3=m(\frac{1}{2}x-\frac{1}{3}y)(\frac{1}{3}y-\frac{1}{4}z)(\frac{1}{4}z-\frac{1}{2}x)$. Find $m$.

Answer 1

${(\frac{1}{2}x-\frac{1}{3}y)}^3+{(\frac{1}{3}y-\frac{1}{4^7})}^3+{(\frac{1}{4^7}-\frac{1}{2}x)}^3=3(\frac{1}{2}x-\frac{1}{3}y).(\frac{1}{3}y-\frac{1}{4^7})(\frac{1}{4^7}-\frac{1}{2}x)$

Let,$\frac{1}{2}x-\frac{1}{3}y=a\frac{1}{3}y-\frac{1}{4^7}=b\frac{1}{4^7}-\frac{1}{2}x=c\therefore a+b+c=\frac{1}{2}x-\frac{1}{3}y+\frac{1}{3}y-\frac{1}{4^7}+\frac{1}{4^7}-\frac{1}{2}x=0\therefore Ifa+b+c=0then{a}^3+b^3+c^3=3abc$

$\therefore$ From above ,we can say that $M=3$.