$lcm(a, b)$ and $lcm(a, c)$ are both divisible by $a$. Taking modulo $a$ on both sides, we see $\gcd(a, b)\equiv\gcd(a, c)\pmod a$. As both are positive and $\le a$, we have that $\gcd(a, b)=\gcd(a, c)$. Thus, $a\cdot b/\gcd(a, b)=lcm(a, b)=lcm(a, c)=a\cdot c/\gcd(a, c)$. Cancelling out $a$ and the gcd functions, we are left with $\boxed{b=c}$.

@all can we do it with Local-Global Principle? Such that $v_p(a) =x$ $v_p(b) =y$ $v_p(c) =z$ Where p is prime Wlog x>=y>=z And we get y=z the can we say b=c(?)

So let's start. First i accepted a=dx b=dy and c=dz. After that gcm(a,b)=d lcm(a,b)=dxy gcm(a,c)=d and finally lcm(a,c)=dxz Let's use it in our equality; d+dxy=d+dxz make it more simple; dxy=dxz so, y=z. İf y is equal to z, also dy=dz ===>b=c We proved)

zaidova wrote: So let's start. First i accepted a=dx b=dy and c=dz. After that gcm(a,b)=d lcm(a,b)=dxy gcm(a,c)=d and finally lcm(a,c)=dxz Let's use it in our equality; d+dxy=d+dxz make it more simple; dxy=dxz so, y=z. İf y is equal to z, also dy=dz ===>b=c I am afraid that I am wrong, but does d make that (x,y) and (x,z)=1? I think you meant the upper according to your solution We proved)

My solution : $gcd(a,b)+lcm(a,b)=gcd(a,c)+lcm(a,c)$ because $a|lcm(a,b),lcm(a,c), gcd(a,b)=gcd(a,c) (mod a) -> gcd(a,b)=gcd(a,c)$ then, $lcm(a,b)=lcm(a,c)$ because$ A*B=gcd*lcm $, $ab=ac$ so, $b=c$