Please help to solve this number theory question based on gcd and lcm. Let $a_1,\ b_1,\ c_1$ be natural numbers.
Let $$\gcd (b_1,c_1)=a_2,\ \gcd(a_1,c_1)=b_2,\ \gcd(a_1,b_1)=c_2$$
$${\rm lcm}(b_2,c_2)=a_3,\ {\rm lcm}(a_2,c_2)=b_3,\ {\rm lcm}(a_2,b_2)=c_3$$
Prove that $$\gcd(b_3,c_3)=a_2$$
I have tried this question a lot but I am stuck at this point.
We have to do something with the power of exponents of primes in $a_1,b_1,c_1$ but can't crack the problem and I am unable to assume the prime factorisation of $a_1,b_1,c_1$.
So please help me to do further and solve the problem.
Thanks
 A: Let the highest exponent of prime $p$ that divides $a_1,b_1,c_1$ respectively be
$$A_1,B_1,C_1$$ 
WLOG $A_1\ge B_1\ge C_1$
So, the highest exponent of $p$ divides $a_2,b_2,c_2,a_3,b_3,c_3$ will respectively be $$C_1,C_1,B_1,B_2,A_2,A_2$$
The highest exponent of $p$ in gcd$(b_3,c_3)=A_2$ which is the same in $c_3$
Now this will hold true for any prime that divides $a_1b_1c_1$
A: Since both $b_3$ and $c_3$ are multiples of $a_2$ by definition, we know that $$a_2 \mid \gcd(b_3,c_3).$$
On the other hand, $b_1$ is (again by definition) a multiple of both $\gcd(b_1,c_1)=a_2$ and $\gcd(a_1,b_1)=c_2$, so therefore $b_1$ is a multiple of lcm$(a_2,c_2) = b_3$. Similarly, $c_1$ is a multiple of both $\gcd(b_1,c_1)=a_2$ and $\gcd(a_1,c_1)=b_2$, so therefore $c_1$ is a multiple of lcm$(a_2,b_2) = c_3$.
Since $b_3\mid b_1$ and $c_3\mid c_1$, it follows that $$\gcd(b_3,c_3) \mid \gcd(b_1,c_1) = a_2.$$
These mutual divisibilities show that $\gcd(b_3,c_3) = a_2$ (since $a_2$ is assumed to be positive).
A: $\DeclareMathOperator{\lcm}{lcm}$
It can be shown using the below properties.
$$\gcd (\lcm(x,y),z)=\lcm (\gcd (x, z), \gcd (y, z))$$
$$\gcd (\lcm (x,y),x)=x$$
$$\gcd (\gcd (x, y), \gcd (z, t)) = \gcd (\gcd (x, z), \gcd (y, t))$$
They are either easy to prove or can be found here. So, I just use them.
$$\gcd (\lcm(a_1,b_1), c_1)=\lcm(\gcd(a_1,c_1),\gcd(b_1,c_1))=\lcm(b_2, a_2)=c_3$$
Similarly,
$$\gcd(\lcm(a_1, c_1), b_1)=b_3.$$
Defining $l_1=\lcm(a_1, b_1)$ and $l_2=\lcm(a_1,c_1)$:
$$\gcd(b_3,c_3)=\gcd(\gcd(l_1,c_1), \gcd (l_2,b_1))=\gcd(\gcd(l_1,b_1), \gcd (l_2,c_1))=\gcd(b_1,c_1)=a_2$$
A: Notation : gcd $\mapsto (\ ),\ {\rm lcm}\ \mapsto ((\ ))$
(1) If $(a_1,b_1,c_1)=T$, then $$ (c_2/T,b_2/T)=1 \ \ast$$ 
Proof : If not, then
for $l>1$ $$ Tl|c_2=(a_1,b_1) ,\ Tl | b_2=(a_1,c_1) $$
Hence $Tl|(a_1,b_1,c_1)$ which is a contradiction
(2) Hence $$b_3=((a_2,c_2))=a_2\frac{c_2}{T},\ c_3=
((a_2,b_2))=a_2\frac{b_2}{T}$$by $\ast$ Hence $ (b_3,c_3)= a_2  $ 
A: Let $gcd(b_1,c_1)=k_1$ , $gcd(c_1,a_1)=k_2$ and $gcd(a_1,b_1)=k_3$ so it will result in $a_2=k_1$ , $b_2=k_2$ and $c_2=k_3$.
Now take $a_3=lcm (c_2,b_2)=lcm(k_3,k_2)=k_3\cdot k_2$ ,
$b_3=lcm(c_2,a_2)=lcm(k_3,k_1)=k_3\cdot k_1$
and $c_3 = lcm(a_2, b_2)=lcm(k_1,k_2)=k_1\cdot k_2$.
Now come to questions $gcd(b_3,c_3)=gcd(k_3\cdot k_1, k_1\cdot k_3)=k_1=a_2$.
Correct me if i am wrong.
