find the value of $p(-\pi)$ and $p(0)$? Let $p(x)$ be a polynomial of degree $7$ with real coefficient such that $p(\pi)=\sqrt3.$ and 
$$\int_{-\pi}^{\pi}x^{k}p(x)dx=0$$
for $0\leq k \leq 6$. Then find the value of $p(-\pi)$ and $p(0)$?
If I take general polynomial then it is difficult to find and also time consuming so please help to solve. Thanks

Consider the space $\mathbb{R}_{\le 7}[x]$ of real polynomials of degree at most $7$, equipped with the inner product $\langle f,g\rangle = \int_{-\pi}^\pi fg$.
Your polynomial $p$ satisfies $p \perp \{1, x, \ldots, x^6\}$. We know that a basis for $\mathbb{R}_{\le 7}[x]$ is given by $\{1, x,\ldots, x^7\}$ so applying Gram-Schmidt process on it gives an orthonormal basis
\begin{equation}
\frac1{\sqrt{2\pi}}, \frac{\sqrt{\frac{3}{2}} x}{\pi ^{3/2}},\frac{3 \sqrt{\frac{5}{2}} \left(x^2-\frac{\pi ^2}{3}\right)}{2 \pi ^{5/2}},\frac{5 \sqrt{\frac{7}{2}} \left(x^3-\frac{3 \pi ^2 x}{5}\right)}{2 \pi ^{7/2}},\\
\frac{3 \left(35 x^4-30 \pi ^2 x^2+3 \pi ^4\right)}{8 \sqrt{2} \pi ^{9/2}},\frac{\sqrt{\frac{11}{2}} \left(63 x^5-70 \pi ^2 x^3+15 \pi ^4 x\right)}{8 \pi ^{11/2}},\\\frac{\sqrt{\frac{13}{2}} \left(231 x^6-315 \pi ^2 x^4+105 \pi ^4 x^2-5 \pi ^6\right)}{16 \pi ^{13/2}},\frac{\sqrt{\frac{15}{2}} \left(429 x^7-693 \pi ^2 x^5+315 \pi ^4 x^3-35 \pi ^6 x\right)}{16 \pi ^{15/2}}
\end{equation}
Hence $p(x) = \frac{\lambda}{16 \pi ^{15/2}}\sqrt{\frac{15}{2}} \left(429 x^7-693 \pi ^2 x^5+315 \pi ^4 x^3-35 \pi ^6 x\right)$ for some constant $\lambda \in \mathbb{R}$. Find $\lambda$ by using the condition $p(\pi) = \sqrt{3}$ and you will have found $p$.
