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Mathematics Question on Determinants

Using properties of determinants,prove that:
\begin{vmatrix} x & x^2 & 1+px^3\\\ y & y^2 & 1+py^3\\\z&z^2&1+pz^3 \end{vmatrix}$$=(1+pxyz)(x-y)(y-z)(z-x)

Answer

Δ=$$\begin{vmatrix} x & x^2 & 1+px^3\\\ y & y^2 & 1+py^3\\\z&z^2&1+pz^3 \end{vmatrix}
Applying R2→R2−R1R_2\rightarrow R_2-R_1 and R3→R3−R1R_3\rightarrow R_3-R_1,we have
=∣xx21+px3 y−xy2−x2p(y3−x3)\z−xz2−x2p(z3−x3)∣\begin{vmatrix} x & x^2 & 1+px^3\\\ y-x & y^2-x^2 & p(y^3-x^3)\\\z-x&z^2-x^2&p(z^3-x^3)\end{vmatrix}
=(y-x)(z-x)$$\begin{vmatrix} x & x^2 & 1+px^3\\\ 1 & y+x & p(y^2+x^2+xy)\\\1&z+x&p(z^2+x^2+xz)\end{vmatrix}

Applying R3→R3−R2R_3\rightarrow R_3-R_2,we have:
Δ=(y-x)(z-x)$$\begin{vmatrix} x & x^2 & 1+px^3\\\ 1& y+x & p(y^2+x^2+xy)\\\0&z-y&p(z-y)(x+y+z)\end{vmatrix}
=(y-x)(z-x)(z-y)$$\begin{vmatrix} x & x^2 & 1+px^3\\\ 1& y+x & p(y^2+x^2+xy)\\\0&1&p(x+y+z)\end{vmatrix}

Expanding along R3R_3,we have:
Δ=(x−y)(y−z)(z−x)[(−1)(p)(xy2+x3+x2y)+1+px3+p(x+y+z)(xy)]Δ=(x-y)(y-z)(z-x)[(-1)(p)(xy^2+x^3+x^2y)+1+px^3+p(x+y+z)(xy)]
=(x−y)(y−z)(z−x)[−pxy2−px3−px2y+1+px3+px2y+pxy2+pxyz]=(x-y)(y-z)(z-x)[-pxy^2-px^3-px^2y+1+px^3+px^2y+pxy^2+pxyz]
=(x−y)(y−z)(z−x)(1+pxyz)=(x-y)(y-z)(z-x)(1+pxyz)

Hence,the given result is proved.