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Solution
$$3^{210} = (3^2)^{105} = 9^{105} < 17^{105} => 3^{210} < 17^{105}$$
$$7^{140} = (7^4)^{35} = 2401^{35}$$
$$17^{105} = (17^3)^{35} = 4913^{35} => 7^{140} < 17^{105}$$
$$17^{105} = (17^5)^{21} = 1419857^{21}$$
$$31^{84} = (31^4)^{21} = 923521^{21} => 31^{84} < 17^{105}$$
Hence, largest of the number = $$17^{105}$$
