Question

ajutor, urgent, dau coroana!

Answer 1

$\\ \det \begin{pmatrix}a&b&c\\ d&e&f\\ g&h&i\end{pmatrix}=a\cdot \det \begin{pmatrix}e&f\\ h&i\end{pmatrix}-b\cdot \det \begin{pmatrix}d&f\\ g&i\end{pmatrix}+c\cdot \det \begin{pmatrix}d&e\\ g&h\end{pmatrix} \\ \\ \\ \det \begin{pmatrix}a^2&\left(a+1\right)^2&\left(a+2\right)^2 \\ b^2&\left(b+1\right)^2&\left(b+2\right)^2 \\ c^2&\left(c+1\right)^2&\left(c+2\right)^2\end{pmatrix} =>$

$\\ a^2 \cdot \det \begin{pmatrix}\left(b+1\right)^2&\left(b+2\right)^2\\ \left(c+1\right)^2&\left(c+2\right)^2\end{pmatrix} - (a+1)^2 \det \begin{pmatrix}b^2&\left(b+2\right)^2\\ c^2&\left(c+2\right)^2\end{pmatrix} \\ + (a+2)^2\det \begin{pmatrix}b^2&\left(b+1\right)^2\\ c^2&\left(c+1\right)^2\end{pmatrix} \\ \\\det \begin{pmatrix}\left(b+1\right)^2&\left(b+2\right)^2\\ \left(c+1\right)^2&\left(c+2\right)^2\end{pmatrix} = \left(b+1\right)^2\left(c+2\right)^2-\left(b+2\right)^2\left(c+1\right)^2 =>$

$=2b^2c+3b^2+4b-2bc^2-3c^2-4c \\ \\ \det \begin{pmatrix}b^2&\left(b+2\right)^2\\ c^2&\left(c+2\right)^2\end{pmatrix} = b^2\left(c+2\right)^2-\left(b+2\right)^2c^2 =4b^2c+4b^2-4bc^2-4c^2 \\ \\ \\ \det \begin{pmatrix}b^2&\left(b+1\right)^2\\ c^2&\left(c+1\right)^2\end{pmatrix} = b^2\left(c+1\right)^2-\left(b+1\right)^2c^2 =2b^2c+b^2-2bc^2-c^2$

$\\ Acum \:\ vine: a^2(2b^2+3b^2+4b-2bc^2-3c^2-4c) - \\ (a+1)^2(4b^2c+4b^2- 4bc^2 -4c^2) + (a+2)^2(2b^2c+b^2-2bc^2-c^2) \\ = 4a^2b-4a^2c-4ab^2+4ac^2+4b^2c-4bc^2 \\ = \boxed{4\left(a^2b-a^2c+ac^2-ab^2+b^2c-bc^2\right)}$