CUET Preparation Today
If p, q, rare distinct, then value of $\begin{vmatrix}p&p^2&1+p^3\\q&q^2&1+q^3\\r&r^2&1+r^3\end{vmatrix}$ is: |
$(1+pqr)(q-p)(r-p)(r-q)$ $(1-pqr)(q+p)(r+p)(r-q)$ $(1+pqr) (q-p)(r+p)(r−q)$ $(1-pqr)(q+p)(r-p)(r+q)$ |
$(1+pqr)(q-p)(r-p)(r-q)$ |
The correct answer is Option (1) → $(1+pqr)(q-p)(r-p)(r-q)$ $\begin{vmatrix}p&p^2&1+p^3\\q&q^2&1+q^3\\r&r^2&1+r^3\end{vmatrix}=\begin{vmatrix}p&p^2&1\\q&q^2&1\\r&r^2&1\end{vmatrix}+\begin{vmatrix}p&p^2&p^3\\q&q^2&q^3\\r&r^2&r^3\end{vmatrix}$ $=\begin{vmatrix}p&p^2&1\\q&q^2&1\\r&r^2&1\end{vmatrix}+pqr\begin{vmatrix}1&p&p^2\\1&q&q^2\\1&r&r^2\end{vmatrix}$ $=(1+pqr)(q-p)(r-p)(r-q)$ |
