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Can anyone tell me the value of the given expression. 😢
a=5,b=2
a=10, b=4.
Let 2^1/3=u. So, u^3=2. x=u^2+u+1= (u^3-1)/(u-1) = 1/(u-1). So, 1/x=u-1. Therefore, 1/x^4+3/x^3+1/x^2+3= u^4-u^3-2u^2+3u+2 = u^4-2u^2+3u= 5u-2u^2. Again, 1/u^2=u/2 and 1/u= u^2/2. So, 5u-2u^2=(a/2)u -(b/2)u^2. So, a=10, b=4.
α=10 β=4
x=³√4+³√2+1 =>x•(³√2-1)=(³√2²+³√2+1²)(³√2-1)=³√2³-1³=2-1=1 => 1/x =³√2-1 =>1/x⁴ +3/x³ +1/x²+3=(1/x)⁴+3(1/x)³+(1/x)²+3=(³√2-1)⁴+3(³√2-1)³+(³√2-1)²+3=5•³√2 -2•³√4 =(5• ³√2•³√4)/³√4-(2•³√4•³√2)/³√2 =5•³√8/³√4-2•³√8/³√2==5•2/³√4-2•2/³√2=10/³√4-4/³√2=> a =10 and b =4.
Terrible manipulation, has to be more systematic way. It's impossible to envisaged this jugglery in exam
Can anyone tell me the value of the given expression. 😢
a=5,b=2
a=10, b=4.
Let 2^1/3=u. So, u^3=2. x=u^2+u+1= (u^3-1)/(u-1) = 1/(u-1). So, 1/x=u-1. Therefore, 1/x^4+3/x^3+1/x^2+3= u^4-u^3-2u^2+3u+2 = u^4-2u^2+3u= 5u-2u^2. Again, 1/u^2=u/2 and 1/u= u^2/2. So, 5u-2u^2=(a/2)u -(b/2)u^2. So, a=10, b=4.
α=10 β=4
x=³√4+³√2+1 =>
x•(³√2-1)=(³√2²+³√2+1²)(³√2-1)=
³√2³-1³=2-1=1 => 1/x =³√2-1 =>
1/x⁴ +3/x³ +1/x²+3=
(1/x)⁴+3(1/x)³+(1/x)²+3=
(³√2-1)⁴+3(³√2-1)³+(³√2-1)²+3=
5•³√2 -2•³√4 =
(5• ³√2•³√4)/³√4-(2•³√4•³√2)/³√2 =5•³√8/³√4-2•³√8/³√2=
=5•2/³√4-2•2/³√2=10/³√4-4/³√2
=> a =10 and b =4.
Terrible manipulation, has to be more systematic way. It's impossible to envisaged this jugglery in exam