Let $p,q,r$ be the roots of $P$. Then $Q$ must satisfy $P\left(\frac{1}{x}\right)=0$, or $Q(x)=bx^3+ax+1=0$, for $x\subset \{\frac{1}{p}, \frac{1}{q},\frac{1}{r}\}$. Pairing coefficients with the given $Q$ we get $a=b=1$. So $P(x)=x^3+x^2+1$ and $Q(x)=x^3+x+1$. Hence \begin{align*}\gcd(P(x),Q(x)) &=\gcd(x^3+x^2+1,x^3+x+1)\\ &=\gcd(3,x-1)\\ &\Rightarrow \gcd(P(2013!+1), Q(2013!+1))\\&=\gcd(3,2013!)\\ &=\boxed{3}.\end{align*}

how is gcd equal to 3,x-1?

How is gcd equal to gcd(3, x-1)?

Yes pls tell

Use euclid's division lemma for polynomials

Since $P(x)-Q(x)=x(x-1)$, we have $\gcd(P,Q)=(P,x(x-1))$. But $P(x)$ and $x$ are coprime and $P(x)=(x-1)(x^2+2x+2)+3$, so $\gcd(P,Q)=\gcd(P,x-1)=\gcd(3,x-1)$.

$P(x)=x^3+ax^2+b$ sps root of this are $\alpha, \beta , \gamma$ , so roots of $Q(x)$ are $\frac{1}{\alpha} , \frac{1}{\beta}, \frac{1}{\gamma}$ we have $ \mathcal{H}(t):=t^3 \cdot P\left(\frac{1}{t}\right)$ roots as $\frac{1}{\alpha}, \frac{1}{\beta} , \frac{1}{\gamma}$ $\mathcal{H}(t)=bt^3+at+1$ , from viet'as relation we have $a=b^2$ , $ab=1 \implies b=a=1 \in \mathbb{Z}$. Now $P(x)=x^3+x^2+1$ , $Q(x)=x^3+x+1$ , we set $t=2013!+1$ ,now $$\displaystyle \gcd(P(2013!+1), Q(2013!+1))= \gcd(t^3+t^2+1, t^3+t+1)=\gcd(t^3+t+1,t^2-t)$$$$ \implies \gcd(t^2+t+1,t^2-t)=\gcd(2t+1,t^2-t)=\gcd(2t+1,3t^2)=\gcd(2t+1,6t^2)=\gcd(2t+1,3t)=\gcd(2t+1,6t)=\gcd(2t+1,3)=3.$$

Let m,n and p be the 3 roots of the equation Q(x) = 0 and 1/m,1/n and 1/p of the equation P(x) = 0. Then, by Vieta's Relation we've mn + np + pm = b (Sum of the roots taken 2 at a time) and, mnp = -a (Product of the roots) Again, from the second equation i.e. p(x) = 0 we have 1/m + 1/n + 1/p = -a (Sum of the roots taken 2 at a time) mn + np + pm/mnp = -a b = a² ————(1) 1/mnp = -b (Product of the roots) ab = 1 On substituting b = a² from equation (1) in the above equation, we get a³ = 1 Therefore, a = b = 1 since a and b are non zero real numbers. So, P(x) = x³ + x² + 1 and Q(x) = x³ + x + 1 Now, gcd[P(x),Q(x)] = gcd(x³ + x² + 1, x³ + x + 1) = gcd(x³ + x + 1,x² - x) = gcd(x² + x + 1,x² - x) = gcd[x(x - 1), 2x + 1] = gcd(x - 1,2x + 1) since gcd(x, 2x + 1) = 1 = gcd(3,x - 1) On plugging x = 2013! + 1 in the above equation, we get gcd[P(2013! + 1, Q(2013! + 1)] = gcd(3, 2013!) = 3 Therefore, the greatest common divisor of P(2013! + 1) and Q(2013! + 1) is 3.