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Question

Question 7
Find the value of 4(216)23+1(256)34+2(243)15

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Solution

We have, 4(216)23+1(256)34+2(243)15
=4(63)23+1(162)34+2(35)15=463×(23)+1162×(34)+2(35)15[(am)n=amn]=462+11632+231=4×62+1632+2×31[1a=a1]=4×36+((4)2)3/2+2×31=4×36+43+6=144+64+6=214

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