@missionCAT13 said:Can any one solve these questions ?1) x^1/3+y^1/3+z^1/3 = 0 then (x+y+z)^3 =
If, (a + b + c) = 0, then (a^3 + b^3 + c^3) = 3abc
=> (x + y + z) = 3(xyz)^1/3
=> (x + y + z)^3 = 27xyz ?
@Logrhythm said:1) x=y=z=02) x=9/4sorry for not posting the explaintion, in office
Sir , 1st me I have specified condition that x is not equal to 0,1 .
@ScareCrow28 said:If, (a + b + c) = 0, then (a^3 + b^3 + c^3) = 3abc=> (x + y + z) = 3(xyz)^1/3=> (x + y + z)^3 = 27xyz ?
thank you sir@missionCAT13 said:Can any one solve these questions ?1) x^1/3+y^1/3+z^1/3 = 0 then (x+y+z)^3 = ?2) x^(x* sqr rt (x)) = (x*sqr rt (x))^x then x = ? (x is not equal to 1,0.)
1. 0
2. 9/4
2. 9/4
@ScareCrow28 said:If, (a + b + c) = 0, then (a^3 + b^3 + c^3) = 3abc=> (x + y + z) = 3(xyz)^1/3=> (x + y + z)^3 = 27xyz ?
another one. :(
1/(x^b + x^-c +1) + 1/(x^c + x^-a +1) + 1/(x^a + x^-b +1) kitna agar a+b+c = 0 ?
@missionCAT13 said:Sir , 1st me I have specified condition that x is not equal to 0,1 .
ok use this
x+y+z=0 then x^3+y^3+z^3 = 3xyz
@missionCAT13 said:another one. 1/(x^b + x^-c +1) + 1/(x^c + x^-a +1) + 1/(x^a + x^-b +1) kitna agar a+b+c = 0 ?
1??
a=b=c=0 again :D
@missionCAT13 said:Yes. But How did you get this ?
it can be written as
(x)^x_/x=(x_/x)^x
=> (x)^x_/x=(x^3/2)^x
=> (x)^x_/x=(x)^3/2x
since x is not equal to 0 or 1.. therefore x_/x=3/2*x
i.e. _/x=3/2 =>x=9/4
(x)^x_/x=(x_/x)^x
=> (x)^x_/x=(x^3/2)^x
=> (x)^x_/x=(x)^3/2x
since x is not equal to 0 or 1.. therefore x_/x=3/2*x
i.e. _/x=3/2 =>x=9/4
@missionCAT13 said:another one. 1/(x^b + x^-c +1) + 1/(x^c + x^-a +1) + 1/(x^a + x^-b +1) kitna agar a+b+c = 0 ?
Ans should be 1
But I am trying to prove it rigorously :P
@missionCAT13 said:another one. 1/(x^b + x^-c +1) + 1/(x^c + x^-a +1) + 1/(x^a + x^-b +1) kitna agar a+b+c = 0 ?
1
@ScareCrow28 said:Ans should be 1But I am trying to prove it rigorously
@Logrhythm said:1??a=b=c=0 again
Yeah, Answer is 1 but all I need is solution for the same.
@missionCAT13 said:Yeah, Answer is 1 but all I need is solution for the same.
a+b+c=0
x^(a+b+c)=1
x^(b+c)=x^(-a)
X^(a+c)=x^(-b)
1/(x^b + x^-c +1)= x^c/(x^(-a) + x^c + 1)----(1)
1/(x^c + x^-a +1)----(2)
1/(x^a + x^-b +1)= x^(-a)/(x^(-a) + x^c + 1)---(3)
Clearly, (1)+(2)+(3)=1
@RDN said:a+b+c=0x^(a+b+c)=1x^(b+c)=x^(-a)X^(a+c)=x^(-b)1/(x^b + x^-c +1)= x^c/(x^(-a) + x^c + 1)----(1)1/(x^c + x^-a +1)----(2)1/(x^a + x^-b +1)= x^(-a)/(x^(-a) + x^c + 1)---(3)Clearly, (1)+(2)+(3)=1
here goes another 1.
Find the value of
1/(1+p+q^-1) + 1/(1+q+r^-1) + 1/(1+r+p^-1)
given that pqr=1.
@missionCAT13 said:here goes another 1.Find the value of 1/(1+p+q^-1) + 1/(1+q+r^-1) + 1/(1+r+p^-1)given that pqr=1.
let p=x^a, q=x^b,, r=x^c for some real a,b,c, x...the above two questions are the same
find 1^2+2^2 +3^2+5^2+8^2+...upto 100 terms.
p.s.- no OA available
p.s.- no OA available
@Joey_Sharma said:find 1^2+2^2 +3^2+5^2+8^2+...upto 100 terms.p.s.- no OA available
We cant simply find the 100th term of a fibonacci series... it should be mentioned unless it is undoable by the conventional formula..
@albiesriram said:We cant simply find the 100th term of a fibonacci series... it should be mentioned unless it is undoable by the conventional formula..
actually this question was posted by u earlier and i couldnt find its solution.. thats why i posted it again
@albiesriram said:oh yes. that was a stupidity by me done in 2012 thread. later understood that we cannot find the 100th term by any means..i wanted to use the recursion and fibonacci concepts. but we can find the 100th term.. Then clarified by @maddy2807 .
actually u posted this question in 2013 quant thread only :P