12)
WLOG,a>b>c
so, a-b>0
b-c>0
c-a<0
that implies, (c-a)1/3 is imaginary whereas other 2 expressions are real.
so,sum of real and imaginary number can never be zero(a,b,c distinct)
1)prove that (2+√5)1/3 + (2-√5)1/3 is rational.
2)Let a,b,c be distinct real numbers.prove that (a-b)1/3+ (b-c)1/3+ (c-a)1/3≠0
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16 Answers
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21Adding one more.
prove the AM-GM inequality for 3 terms ie x3 + y3 + z3 ≥3xyz
1@#6
a3 + b3 + c3 = (a + b + c) (a2 + b2 + c2 – ab – bc – ca) + 3abc
11let 3√(a-b)=x,3√(b-c)=y,3√(c-a)=z.
then,x3+y3+z3=0
either,(x+y+z)=0
or,x=y=z...[then,it will be a=b=c]
oh...my god..i have proved..
3√(a-b)+3√(b-c)+3√(c-a)=0
11let,(2+√5)1/3=a,2+√5=a3
(2-√5)1/3=b,2-√5=b3
a3+b3=0
(a+b)3-3ab(a+b)=0...
plz..help me from here..
21@soham how u got that if x3 + y3 + z3 =0
either x+ y+z =0 or x=y=z ?? this is not correct
21@ #10 you have a3 + b3 = 4 (not zero)
so (a+b)3 - 3ab(a+b) = 4
or (a+b)3 + 3(a+b) = 4 (ab= -1)
solving the equation for (a+b) we will get (a+b) = 1 hence its proved that the expression is rational.
it can also be solved this way
let x= (2+√5)1/3 + (2-√5)1/3
so x- (2+√5)1/3 - (2-√5)1/3=0
so x3 - 2- √5 - 2 + √5= 3x (2+√5)1/3 (2-√5)1/3
(using the concept that if a+b+c = 0, a3 + b3 +c3 =3abc)
or x3 - 4 = - 3x
or x=1
there is no other real root.
so the expression is a rational number.
36we know that cuberoot(-ve term) is imaginary...!!
so, how did u take ab = -1 ???
Again,
in ur soln...
i didn't get how u used the concept of a3 + b3 + c3 = 3abc, when a + b + c = 0
Please explain....!! i didn't get..!!
11no,3√of -ve is not imaginary...
like3√-1=-1
other two roots may be ignored...
36@Soham -> Thanks for clearing my doubt
Well your solution was wonderful and easy to understand....!!
Thanks..!!