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1) ![\dfrac{ace}{bdf}=k^3\to ace = (bdf)k^3 \dfrac{ace}{bdf}=k^3\to ace = (bdf)k^3](https://tex.z-dn.net/?f=%5Cdfrac%7Bace%7D%7Bbdf%7D%3Dk%5E3%5Cto+ace+%3D+%28bdf%29k%5E3)
2)
![\dfrac{a+b}{b}=\dfrac{c+d}{d}=\dfrac{e+f}{f}=k+1\to (a+b)(c+d)(e+f)=bdf(k+1)^3\\ \\
bdf(k+1)^3=8^{16}=2^{48}\to bdf=\dfrac{2^{48}}{(k+1)^3}\to ace =\dfrac{2^{48}k^3}{(k+1)^3}\\ \\ \\ \\
\sqrt[3]{ace}+\sqrt[3]{bdf}=\sqrt[3]{\dfrac{2^{48}k^3}{(k+1)^3}}+\sqrt[3]{\dfrac{2^{48}}{(k+1)^3}}=\dfrac{2^{16}(k+1)}{k+1}=2^{16} \dfrac{a+b}{b}=\dfrac{c+d}{d}=\dfrac{e+f}{f}=k+1\to (a+b)(c+d)(e+f)=bdf(k+1)^3\\ \\
bdf(k+1)^3=8^{16}=2^{48}\to bdf=\dfrac{2^{48}}{(k+1)^3}\to ace =\dfrac{2^{48}k^3}{(k+1)^3}\\ \\ \\ \\
\sqrt[3]{ace}+\sqrt[3]{bdf}=\sqrt[3]{\dfrac{2^{48}k^3}{(k+1)^3}}+\sqrt[3]{\dfrac{2^{48}}{(k+1)^3}}=\dfrac{2^{16}(k+1)}{k+1}=2^{16}](https://tex.z-dn.net/?f=%5Cdfrac%7Ba%2Bb%7D%7Bb%7D%3D%5Cdfrac%7Bc%2Bd%7D%7Bd%7D%3D%5Cdfrac%7Be%2Bf%7D%7Bf%7D%3Dk%2B1%5Cto+%28a%2Bb%29%28c%2Bd%29%28e%2Bf%29%3Dbdf%28k%2B1%29%5E3%5C%5C+%5C%5C%0Abdf%28k%2B1%29%5E3%3D8%5E%7B16%7D%3D2%5E%7B48%7D%5Cto+bdf%3D%5Cdfrac%7B2%5E%7B48%7D%7D%7B%28k%2B1%29%5E3%7D%5Cto+ace+%3D%5Cdfrac%7B2%5E%7B48%7Dk%5E3%7D%7B%28k%2B1%29%5E3%7D%5C%5C+%5C%5C+%5C%5C+%5C%5C%0A%5Csqrt%5B3%5D%7Bace%7D%2B%5Csqrt%5B3%5D%7Bbdf%7D%3D%5Csqrt%5B3%5D%7B%5Cdfrac%7B2%5E%7B48%7Dk%5E3%7D%7B%28k%2B1%29%5E3%7D%7D%2B%5Csqrt%5B3%5D%7B%5Cdfrac%7B2%5E%7B48%7D%7D%7B%28k%2B1%29%5E3%7D%7D%3D%5Cdfrac%7B2%5E%7B16%7D%28k%2B1%29%7D%7Bk%2B1%7D%3D2%5E%7B16%7D)
2)
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