$\triangle \left( x \right) =\begin{vmatrix} 2{ x }^{ 3 }-3{ x }^{ 2 } & 5x+7 & 2 \\ 4{ x }^{ 3 }-7x & 3x+2 & 1 \\ 7{ x }^{ 3 }-8{ x }^{ 2 } & x-1 & 3 \end{vmatrix}={ a }_{ 0 }+{ a }_{ 1 }x+....{ a }_{ 4 }{ x }^{ 4 }$

$\displaystyle a_{0}$ equals

If the points $(2,5),(4,6)$ and $(a,a)$ are collinear, then the value of $a$ is equal to

Three digits numbers $7x,36y$ and $12z$ where $x , y , z$ are integers from $0$ to $9 ,$ are divisible by a fixed constant $k.$ Then the determinant $\left| \begin{array} { l l l } { x } & { 3 } & { 1 } \\ { 7 } & { 6 } & { z } \\ { 1 } & { y } & { 2 } \end{array} \right|$ $\ +48$ must be divisible by

$\mathrm{K}$ এর কোন মানের জন্য $\left[\begin{array}{cc}K+1 & 3 \\ 3 & K-1\end{array}\right]$ ম্যাট্রিক্সটি বিপরীতযোগ্য নয়?

lf the lines $3\mathrm{x}+2\mathrm{y}-5=0,\ 2\mathrm{x}-5\mathrm{y}+3=0,\ 5\mathrm{x}+\mathrm{b}\mathrm{y}+\mathrm{c}=0$ are concurrent then $\mathrm{b}+\mathrm{c}=$