$\left| {\begin{array}{*{20}{c}}{{b^2} + {c^2}}&{{a^2}}&{{a^2}}\\{{b^2}}&{{c^2} + {a^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{a^2} + {b^2}}\end{array}} \right| = $

The value of the determinant $\left| \begin{array}{ccc} a & a+b & a+2b \\ a+2b & a & a+b \\ a+b & a+2b & a \end{array} \right|$ is

If $\Delta _1 = \left| \begin{matrix} b^5c^6(c^3 - b^3) & a^4c^6(a^3 - c^3) & a^4b^5(b^3 - a^3) \\ b^2c^3(b^6 - c^6) & ac^3(c^6 - a^6) & ab^2(a^6 - b^6) \\ b^2c^3(c^3 - b^3) & ac^3(a^3 - c^3) & ab^2(b^3 - a^3) \end{matrix} \right|$ and $\Delta _2 = \left| \begin{matrix} a & b^2 & c^3 \\ a^4 & b^5 & c^6 \\ a^7 & b^8 & c^9 \end{matrix} \right|$,then $\Delta _1 \Delta _2$ is equal to:

The value of $\left| \begin{array}{ccc} 1 & x & y \\ 2 & \sin x + 2x & \sin y + 2y \\ 3 & \cos x + 3x & \cos y + 3y \end{array} \right|$ is

The value of the determinant $\left| \begin{array}{ccc} 1 & 1 & 1 \\ b+c & c+a & a+b \\ b+c-a & c+a-b & a+b-c \end{array} \right|$ is

Using the property of determinants and without expanding,prove that:
$\left|\begin{array}{lll}1 & bc & a(b+c) \\ 1 & ca & b(c+a) \\ 1 & ab & c(a+b)\end{array}\right|=0$