যদি \((b + c - a)x = (c + a - b)y\) \( = (a + b - c)z = 2\) হয়, তবে দেখাও যে, \(\left(\cfrac{1}{x}+\cfrac{1}{y}\right)\left(\cfrac{1}{y}+\cfrac{1}{z}\right)\left(\cfrac{1}{z}+\cfrac{1}{x}\right)=abc\) Madhyamik 2017


\((b + c - a)x = (c + a - b)y\) \( = (a + b - c)z = 2\)
\(\therefore \cfrac{1}{(b + c - a)x} = \cfrac{1}{(c + a - b)y}\)

\( = \cfrac{1}{(a + b - c)z} = \cfrac{1}{2}\)




\(\therefore \cfrac{1}{x}=\cfrac{b+c-a}{2}\)
\(\cfrac{1}{y}=\cfrac{c+a-b}{2}\)
এবং \(\cfrac{1}{z}=\cfrac{a+b-c}{2}\)

\(\therefore \cfrac{1}{x}+\cfrac{1}{y}=\cfrac{b+c-a}{2}+\cfrac{c+a-b}{2}\)
\(=\cfrac{b+c-a+c+a-b}{2}=\cfrac{2c}{2}=c\)

\(\cfrac{1}{y}+\cfrac{1}{z}=\cfrac{c+a-b}{2}+\cfrac{a+b-c}{2}\)
\(=\cfrac{c+a-b+a+b-c}{2}=\cfrac{2a}{2}=a\)

এবং \(\cfrac{1}{z}+\cfrac{1}{x}=\cfrac{a+b-c}{2}+\cfrac{b+c-a}{2}\)
\(=\cfrac{a+b-c+b+c-a}{2}=\cfrac{2b}{2}=b\)

\(\therefore \left(\cfrac{1}{x}+\cfrac{1}{y}\right)\left(\cfrac{1}{y}+\cfrac{1}{z}\right)\left(\cfrac{1}{z}+\cfrac{1}{x}\right)=abc\)

(প্রমাণিত)


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