For a natural number n, let α _n=19^n-12^n. Then, the value of 31α _9-α _10 57α _8 is

If ${r}_{1}$r1​ and ${r}_{2}$r2​ are roots of the quadratic $a{x}^{2}+bx+c=0$ax2+bx+c=0

then ${{r}_{1}}^{n}=31{{r}_{1}}^{n-2}-228{{r}_{1}}^{n-1}$r1​n=31r1​n−2−228r1​n−1 and ${{r}_{2}}^{n}=31{{r}_{2}}^{n-2}-228{{r}_{2}}^{n-1}$r2​n=31r2​n−2−228r2​n−1

So, ${{r}_{1}}^{n}-{{r}_{2}}^{n}=31({{r}_{1}}^{n-2}-{{r}_{2}}^{n-2})-228({{r}_{1}}^{n-1}-{{r}_{2}}^{n-1})$r1​n−r2​n=31(r1​n−2−r2​n−2)−228(r1​n−1−r2​n−1)

Or ${a}_{n}-31{a}_{n-1}+228{a}_{n-2}=0$an​−31an−1​+228an−2​=0 where ${a}_{n}={{r}_{1}}^{n}-{{r}_{2}}^{n}$an​=r1​n−r2​n

Since 19&12 are the roots of the quadratic ${x}^{2}-31x+228=0$x2−31x+228=0

we know, ${a}_{n}-31{a}_{n-1}+228{a}_{n-2}=0$an​−31an−1​+228an−2​=0

Now, for $n=10,{\alpha }_{10}-31{\alpha }_{3}+228{\alpha }_{8}=0$n=10,α10​−31α3​+228α8​=0

$$\Rightarrow \frac{31{\alpha }_{9}-{\alpha }_{10}}{57{\alpha }_{8}}=4$$⇒57α8​31α9​−α10​​=4