Evaluate:
(2−1+3−1)3(2^{-1} + 3^{-1})^3(2−1+3−1)3
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As we know, for any non-zero rational number a
a−n=1ana^{-n} = \dfrac{1}{a^n}a−n=an1 and an=1a−na^{n} = \dfrac{1}{a^{-n}}an=a−n1.
Hence,
(2−1+3−1)3=(121+131)3=(12+13)3(2^{-1} + 3^{-1})^3\\[1em] = \Big(\dfrac{1}{2}^1 + \dfrac{1}{3}^1\Big)^3\\[1em] = \Big(\dfrac{1}{2} + \dfrac{1}{3}\Big)^3(2−1+3−1)3=(211+311)3=(21+31)3
LCM of 2 and 3 is 2 x 3 = 6
=(1×32×3+1×23×2)3=(36+26)3=(3+26)3=(56)3=(5×5×56×6×6)=125216= \Big(\dfrac{1 \times 3}{2 \times 3} + \dfrac{1 \times 2}{3 \times 2}\Big)^3\\[1em] = \Big(\dfrac{3}{6} + \dfrac{2}{6}\Big)^3\\[1em] = \Big(\dfrac{3 + 2}{6}\Big)^3\\[1em] = \Big(\dfrac{5}{6}\Big)^3\\[1em] = \Big(\dfrac{5 \times 5 \times 5}{6 \times 6 \times 6}\Big)\\[1em] = \dfrac{125}{216}=(2×31×3+3×21×2)3=(63+62)3=(63+2)3=(65)3=(6×6×65×5×5)=216125
Hence,(2−1+3−1)3=125216(2^{-1} + 3^{-1})^3 = \dfrac{125}{216}(2−1+3−1)3=216125
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