If e1=((2^1/3)-1)^1/3 e2=(9^-1/3)-((2/9)^1/3)+((4/9)^1/3) then prove that e1=e2.

Question

SHAIK AASIF AHAMED · Answer

Hello student, given [e 1 =(2^{1/3}-1)^{1/3}] so cubing on both sides we get e 1 3 = [(2^{1/3}-1)] ..........(1) Similarly [e 2 =(9^{-1/3})(1-2^{1/3}+2^{2/3})] since [(1-2^{1/3}+2^{2/3})] is in the form of a 2 -ab+b 2 =(a 3 +b 3 )/(a+b) where a=2^1/3and b=1 so [e 2 =3^{1/3}/(2^{1/3}+1)] also cubing above equation on both sides we get e 2 3 = [3/(3+3*2^{1/3}(2^{1/3}+1))] = [1/(1+2^{1/3}+2^{2/3})] as denominator is in the form ofa 2 +ab+b 2 =(a 3 -b 3 )/(a-b) e23= [(2^{1/3}-1)/2-1] &hellip;...............(2) so from 1 and 2 we can see that e1=e2

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If e1=((2^1/3)-1)^1/3 e2=(9^-1/3)-((2/9)^1/3)+((4/9)^1/3) then prove that e1=e2.

If e1=((2^1/3)-1)^1/3 e2=(9^-1/3)-((2/9)^1/3)+((4/9)^1/3) then prove that e1=e2.

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