[ GCD and LCM. Mutual simplicity ]

Find such linear functions $P(x)$ and $Q(x)$ that the equality $P(x)\left(2 x^{3}-7 x^{2}+7 x-2\right)+Q(x)$ $\left(2 x^{3}+x^{2}+x-1\right)=2 x-1$ holds.
Note that $2 x^{3}-7 x^{2}+7 x-2=(x-1)(x-2)(2 x-1), 2 x^{3}+x^{2}+x-1=(2 x-1)\left(x^{2}+x+1\right)$. Therefore, it is sufficient to find such functions $P(x)$ and $Q(x)$ so that the equality $P(x)\left(x^{2}-3 x+2\right)+Q(x)\left(x^{2}+x+1\right)=1$ holds. They can be sought, for example, by the method of undetermined coefficients, writing the last equality in the form $(a x + b)\left(x^{2}-3 x+2\right)-(a x+c)\left(x^{2}+x+1\right)=1$.

## Answer

$P(x)=\frac{1}{21}(4 x+5), Q(x)=-\frac{1}{21}(4 x-11)$.

Simplify the expression: $\frac{(a+b+c)^{5}-a^{5}-b^{5}-c^{5}}{(a+b+c)^{3}-a^{3}-b^{3}-c^{3}}$.

## Solution

$(a+b+c)^{3}-a^{3}-b^{3}-c^{3}=3(a+b)(a+c)(b+c)$ (see problem $\underline{61005}$). Factorize the numerator.

First method. $(a+b)^{5}-a^{5}-b^{5}=5 a b\left(a^{3}+b^{3}\right)+10 a^{2} b^{2}(a+b)=5 a b(a+b)\left(a^{2}+a b+b^{2}\right)$. Therefore, $(a+b+c)^{5}-a^{5}-b^{5}-c^{5}=(a+b+c)^{5}-(a+b)^{5}-c^{5}+(a+b)^{5}-a^{5}-b^{5}=$ $=5 c(a+b)(a+b+c)\left((a+b)^{2}+(a+b) c+c^{2}\right)+5 a b(a+b)\left(a^{2}+a b+b^{2}\right)=$ $=5(a+b)\left(c(a+b+c)\left(a^{2}+b^{2}+c^{2}+2 a b+a c+b c\right)+a b\left(a^{2}+a b+b^{2}\right)\right)=$ $=5(a+b)\left(c(a+b+c)\left(a^{2}+b^{2}+c^{2}+a b+a c+b c\right)+a b c(a+b+c)+a b\left(a^{2}+a b+b^{2}\right)\right)=$ $=5(a+b)\left(a^{2}+b^{2}+c^{2}+a b+a c+b c\right) \cdot(c(a+b+c)+a b)=5(a+b)(b+c)(a+c)\left(a^{2}+b^{2}+c^{2}+a b+a c+b c\right)$.

Second method. By the Remainder Theorem (see problem $\underline{60961}$), the polynomial in the numerator is divisible by $a+b, a+c$, and $b+c$, and therefore by their product. In the quotient, a symmetric polynomial of the second degree should be obtained, that is, of the form $p\left(a^{2}+b^{2}+c^{2}\right)+q(a b+a c+b c)$. Substituting $(a, b, c)=(1,1,0)$ and $(2,-1,0)$, we find that $p=q=5$. (One can also compare the coefficients of $a^{4} b$ and $a^{3} b^{2}$ in the equality $(a+b+c)^{5}-a^{5}-b^{5}-c^{5}=(a+b)(b+c)(a+c)\left(p\left(a^{2}+b^{2}+c^{2}\right)+q(a b+a c+b c)\right)$.)

Thus, $\frac{(a+b+c)^{5}-a^{5}-b^{5}-c^{5}}{(a+b+c)^{3}-a^{3}-b^{3}-c^{3}}=\frac{5\left(a^{2}+b^{2}+c^{2}+a b+a c+b c\right)}{3}$.

## Answer

$\frac{5\left(a^{2}+b^{2}+c^{2}+a b+a c+b c\right)}{3}$.