A Pre-RMO question! -6

Let $P(y)=ay^3+by^2+cy+d$ $P(0)=\boxed{d=k}$ $P(1)=a+b+c+d=a+b+c\cancel{+k}=\cancel{2}k$ $\boxed{a+b+c=k}$ $P(-1)=-a+b-c+d=b-a-c+k=3k$ $b-a-c=2k$ $b=2k+(a+c)=2k+(k-b)=3k-b$ $\boxed{2b=3k}$ Evaluating the expression $P(2)+P(-2)$ $2^3a+2^2b\cancel{+2c}+d+(-2)^3a+2^2b\cancel{-2c}+d=\cancel{8a}+4b+k\cancel{-8a}+4b+k=8b+2k=4(2b)+2k=4(3k)+2k=12k+2k=\boxed{14}k$

Let $P(x) = ax^3 + bx^2 + cx + d$ . Then

$\begin{aligned} P(0) & = d = k \\ P(1) & = a+b+c+k = 2k \\ P(-1) & = -a +b-c +k = 3k \\ \implies P(1)+P(-1) & = 2b + 2k = 5k & \small \blue{\implies 2b = 3k} \\ P(2) & = 8a+4b+2c+k \\ P(-2) & = -8a + 4b - 2c + k \\ \implies P(2) + P(-2) & = 8b + 2k \\ & = 4(2b) + 2k \\ & = 12k + 2k \\ & = 14k \end{aligned}$

Therefore, $\alpha = \boxed{14}$ .