Knowing about Cubic Polynomial

Let ${p(x)} = ax^3+bx^2+cx+d.$

We know that ,

${p(1) = a + b + c + d =1}$ $.............................. \boxed{1}$

${p(2) = 8a + 4b + 2c + d = 2}$ $.......................... \boxed{2}$

${p(3) = 27a + 9b + 3c + d = 3}$ $......................... \boxed{3}$

${p(4) = 64a + 16b + 4c + d = 5}$ $........................ \boxed{4}$ .

By solving these four equations , we get the following values :

$a = \dfrac{1}{6} , b = -1 , c = \dfrac{17}{6} , d = -1$

Therefore ${p(x)} = \dfrac{x^3}{6} - x^2 + \dfrac{17x}{6} - 1.$

By plugging in 6, we get ,

${p(6)} = \dfrac{216}{6} - 36 + 17 - 1 = 36 - 36 + 17 - 1 = 16.$

Therefore the required answer is $16.$

For $x=1,2,3$ we see that $p(x)=x$ , hence $p(x)=a(x-1)(x-2)(x-3)+x$ for some constant $a$ . Also, we are given the value of $p(4)$ , so, by letting $x=4$ : $5=a(4-1)(4-2)(4-3)+4 \implies a=\dfrac{1}{6}$ . So, $p(x)=\dfrac{1}{6}(x-1)(x-2)(x-3)+x$ . Finally, let $x=6$ : $p(6)=\dfrac{1}{6}(6-1)(6-2)(6-3)+6=\boxed{16}$ .

By Lagrange Interpolation formula ,

$P(x)=1×-\frac{1}{6}(x-2)(x-3)(x-4)+2×\frac{1}{2}(x-1)(x-3)(x-4)+3×-\frac{1}{2}(x-1)(x-2)(x-4)+5×\frac{1}{6}(x-1)(x-2)(x-3)$

Plugging the value of $x\rightarrow 6$ ,

$P(6)=16$