Prove that :
(i)(√3×5−3÷3√3−1√5)×6√3×56=35(ii)932−3×50−(181)−12=15(iii)(14)−2−3×823×40+(916)−12=163(iv)212×313×41410−15×535÷343×5−754−35×6=10(v)√14+(0.01)−12−(27)23=32(vi)2n+2n−12n+1−2n=32(vii)(64125)−23+1(256625)14+(√253√64)0=6116(viii)3−3×62×√9852×3√125×(15)−43×313=28√2(ix)(0.6)0−(0.1)−1(38)−1(32)3+(13)−1=−32
(i)√3×5−3÷3√3−1√5×6√3×56=35L.H.S=√3×5−3÷3√3−1√5×6√3×56=(3×5−3)12÷(3−1)13(5)12×(3×56)16=(312×5−32)÷(3−13×512)×(316×56×16)=(312×5−32)÷(3−13×512)×316×51=312−(−13)+16.5−32−12+1=312+13+16.5−3−1+22=33+2+16.5−4+22=366×5−22=31.5−1=35=R.H.S
(ii)932−3×50−(181)−12=15L.H.S=932−3×50−(181)−12=(32)32−3×50−(192)−12 {∵a0=1}=32×32−3−(19)2×(−12)=33−3−(19)−127−3−(19)−127−3−9=15=R.H.S.
(iii)(14)−2−3×823×40+(916)−12=163L.H.S=(14)−2−3×823×40+(916)−12=[(12)2]−2−3×(23)23×1+(34)2×−12=122×−12−3×23×23×1+(34)2×(−12)=12−4−3×22×1+(34)−1=(21)4−3×4+42 {∵a−m=1am}=(21)4−12+43=16−12+43=4+43=163=R.H.S
(iv)212×313×41410−15×535÷343×5−754−35×6=10L.H.S.=212×313×41410−15×535÷343×5−754−35×6=212×313×(22)14(2×5)−15×535÷343×5−75(22)−35×21×31=212×313×22×142−15×5−15×535÷343×5−752−65×21×31=212×313×22×142−15×5−15×535×2−65×21×31343×5−75=212+12−65+1+15.313+1−43.515−35+75==25+5−12+10+210.31+3−43.51−3+75=222−1210.34−4.58−35=21010.30.555=21.30.51=2×1×5=10=R.H.S.
(v)√14+(0.01)−12−(27)23=32L.H.S.=√14+(0.01)−12−(27)23=(122)12+(0.1)2×(−12)−(33)23=1(22)12+(0.1)2×−12−33×23=121+(0.1)−1−32=12+(101)1−32=12+10−9=112=32=R.H.S.
(vi)2n+2n−12n−1−2n=32L.H.S=2n+2n−12n−1−2n=2n+2n×2−12n.21−2n=2n(1+2−1)2n(21−1)=1+122−1=321=32=R.H.S.
(vii)(64125)−23+1(256625)14+(√2503√64)0=6116L.H.S.=(64125)−23+1(256625)14+(√253√64)=(4353)−23+1(4454)14+(√523√43)=43×(−23)53×−23+144×1454×14+(52)12(43)13=5242+54+54=2516+104=2516+4016=6516=R.H.S.
(viii)3−3×62×√9852×3√125×(15)−43×313=28√2L.H.S.=3−3×62×√9852×3√125×(15)−43×313=3−3×(22×32)×(2×49)1252×(25)−13×3−43×5−43×313=3−3×22×32×212×(72)1252×(52)−13×3−43×5−43×313=22.212.3−3+2+43−13.523−2+43.71=4√2×30.50×71=4√2×1×1×7=28√2=R.H.S.
(ix)(0.6)0−(0.1)−1(38)−1(32)3+(−13)−1=−32L.H.S=(0.6)0−(0.1)−1(38)−1(32)3+(−13)−1=1−10138×278+(−31)=−99−31=−96=−32=RHS