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]=46−2+116−32+23−1=4×62+1632+2×31[∵1a=a−1]=4×36+((4)2)3/2+2×31=4×36+43+6=144+64+6=214