Prove the following:
x+y+zx−1y−1+y−1z−1+z−1x−1=xyz\dfrac{x + y + z}{x^{-1}y^{-1} + y^{-1}z^{-1} + z^{-1}x^{-1}} = xyzx−1y−1+y−1z−1+z−1x−1x+y+z=xyz
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Given,
Solving L.H.S. of above equation,
⇒x+y+z1xy+1yz+1zx⇒x+y+zz+x+yxyz⇒xyz(x+y+z)x+y+z⇒xyz.\Rightarrow \dfrac{x + y + z}{\dfrac{1}{xy} + \dfrac{1}{yz} + \dfrac{1}{zx}} \\[1em] \Rightarrow \dfrac{x + y + z}{\dfrac{z + x + y}{xyz}} \\[1em] \Rightarrow \dfrac{xyz(x + y + z)}{x + y + z} \\[1em] \Rightarrow xyz.⇒xy1+yz1+zx1x+y+z⇒xyzz+x+yx+y+z⇒x+y+zxyz(x+y+z)⇒xyz.
Hence, proved that x+y+zx−1y−1+y−1z−1+z−1x−1\dfrac{x + y + z}{x^{-1}y^{-1} + y^{-1}z^{-1} + z^{-1}x^{-1}}x−1y−1+y−1z−1+z−1x−1x+y+z = xyz.
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Simplify the following:
11+am−n+11+an−m\dfrac{1}{1 + a^{m - n}} + \dfrac{1}{1 + a^{n - m}}1+am−n1+1+an−m1
(a + b)-1(a-1 + b-1) = 1ab\dfrac{1}{ab}ab1
If a = cz, b = ax and c = by, prove that xyz = 1.
If a = xyp - 1, b = xyq - 1 and c = xyr - 1, prove that
aq - r.br - p.cp - q = 1.