Isko bhi solve kardijiye

Ligiye sir समझ ligiye जब आपको olampiyad level के questions hi solve करने hain to

Plzzz solution...

Ap
b=a+c/(2)

Gp
c²=ad

Hp
d=ce/(c+e)

c²/a=ce/(c+e)
c/a=e/(c+e)
c²+ce=ea
c²+ce-ea=0

c=-e±√e²+4e/(2)
2c=-e+√e²+4e
(2c+e)²=(e²+4e)
4c²+e²+2ec=e²+4e
2c²+ec=2e
😐

Let x+y= a , y+z= b, z+x= c

Now,

Given, a⅓+b⅓+c⅓ = 0
If a+b+c=0 then a³+b³+c³= 3abc
So,
a+b+c = 3a⅓b⅓c⅓

a+b+c = 3(abc)⅓

2(x+y+z) = 3[(x+y)(y+z)(z+x)]⅓

8[x³+y³+z³+3(x+y)(y+z)(z+x)] = 27(x+y)(y+z)(z+x)

=> 8(x³+y³+z³) = 3(x+y)(y+z)(z+x)

=> (x³+y³+z³)= 3/8(x+y)(y+z)(z+x)

??

Put
bx + ay = ab/(a²+b²)+ ab/(a²+b²)

x = a/(a² + b²)
y = b/(a²+b²)

a² + b²/(a²+b²)² * a² + b²

=(a² + b²)/(a² + b²)

= 1

x^2+x=5
x^2+6x+9-5x-14=0
(x+3)^2-5(x+3)+1=0
(x+3)^2+1=5(x+3)

=>
x+3+1/(x+3)=5
x+3=a
a^3+1/a^3=(a+1/a)^3-3(a+1/a)=
5^3-3•5=110

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arctgx+arcsinx=pi/2
sin(arctgx+arcsinx)=sinpi/2

arcsinx=arctg(tgarcsinx)=
arctg(sinarcsinx/cosarcsinx)=
artg(x/sqrt(1-x^2))=
pi/2-arcctg(x/sqrt(1-x^2))

arctgx=pi/2-arcctgx

pi/2-arcctgx+pi/2-arcctg(x/sqrt(1-x^2))=pi/2
arctg(x/sqrt(1-x^2))+arcctgx=pi/2

Ikkala tarafini ctg lavorsak:

ctg(arctgx/sqrt(1-x^2)+arctgx)=ctgpi/2

(x/sqrt[1-x^2]•x-1)/(x/sqrt[1-x^2]+x)=0
x^2=sqrt(1-x^2)
Va x(1+sqrt[1-x^2]) teng emas 0

x^4=1-x^2
x^4+x^2-1=0
D=1+4=5
x^2=(-1+✔️5)/2
x=sqrt(-1+✔️5)/2

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https://t.me/joinchat/AAAAAEq1X9CmyP2a45xplA

Pls Give solution👏👏👏

Put c=0

a+b=1
a^3+b^3=4

a^3+b^3=(a+b)^3-3ab(a+b)
4=1-3ab
3ab=-3
ab=-1

1/a+1/b+1/ab=(a+b+1)/ab=(1+1)/-1=-2