Question

[tex] \left |\begin{array}{ccc} 1&x&x³ \\1 &y&y³ \\ 1&z&z ³\end{array}\right | =(x-y)(y-z)(z-x)(x+y+z),prove it using theorems [\tex]

Answer 1

Step-by-step explanation:

$\left|</p><p>\begin {array}{ccc}</p><p> 1& x& x^3 \\</p><p>1 & y& y^3 \\</p><p> 1& z& z ^3\end{array}</p><p>\right|$$=(x-y)(y-z)(z-x)(x+y+z)$

To prove it using theorems,

R1-> R1-R2

R2-> R2-R3

$\left|</p><p>\begin {array}{ccc}</p><p> 1-1& x-y& x^3-y^3\\</p><p>1-1 & y-z& y^3-z ^3 \\</p><p> 1& z& z ^3\end{array}</p><p>\right|$

$\left|</p><p>\begin {array}{ccc}</p><p> 0& x-y& (x-y)(x^2+xy+y^2)\\</p><p>0 & y-z& (y-z)(y^2+yz+z ^2 \\</p><p> 1& z& z ^3\end{array}</p><p>\right|$

take common (x-y) from R1 and (y-z) from R2

$(x-y)(y-z) \left|</p><p>\begin {array}{ccc}</p><p> 0& 1& (x^2+xy+y^2)\\</p><p>0 & 1& (y^2+yz+z ^2)\\</p><p> 1& z& z ^3\end{array}</p><p>\right|$

R1->R1-R2

$(x-y)(y-z) \left|</p><p>\begin {array}{ccc}</p><p> 0& 1-1& (x^2+xy+y^2)-(y^2+yz+z ^2)\\</p><p>0 & 1& (y^2+yz+z ^2) \\</p><p> 1& z& z ^3\end{array}</p><p>\right|$

$(x-y)(y-z)\left|</p><p>\begin {array}{ccc}</p><p> 0& 0& (x^2+xy-yz-z ^2)\\</p><p>0 & 1& (y^2+yz+z ^2) \\</p><p> 1& z& z ^3\end{array}</p><p>\right|$

Expand the determinant along column C1

$(x^2+xy-yz-z ^2)\\\\x^2-z^2+xy-yz\\\\(x+z)(x-z)+y(x-z)\\\\(x-z)(x+y+z)$

Now multiply this to the terms taken out common earlier

$(x-y)(y-z)(z-x)(x+y+z)$

= RHS

hence proved

Answer 2

Answer:

I have applied factor theorem to prove this problem

$Let\:\triangle=\left|\begin{array}{ccc}1&x&x^3\\1&y&y^3\\1&z&z^3\end{array}\right|$

put x=y

$Let\:\triangle=\left|\begin{array}{ccc}1&y&y^3\\1&y&y^3\\1&z&z^3\end{array}\right|$

$\text{since }R_1\text{ and }R_2\text{ are identical, }\triangle=0$

$\therefore\text{ (x-y) is a factor of }\triangle$

$\text{similarly, (y-z) and (z-x) are factors of }\triangle$

$\text{Hence, (x-y)(y-z)(z-x) is a factor of }\triangle$

$\text{Its degree is 3}$

$\text{Product of leading diagonal elements = }1.y.z^3$

$\text{Its degree is 4}$

$\text{Difference, m=4-3=1}$

By symmetric and cyclic property, the remaining factor must be k(x+y+z)

$\implies\:\triangle=\left|\begin{array}{ccc}1&x&x^3\\1&y&y^3\\1&z&z^3\end{array}\right|=k(x-y)(y-z)(z-x)(x+y+z)$

put x=0, y=1 and z=2

$\triangle=\left|\begin{array}{ccc}1&0&0\\1&1&1\\1&2&8\end{array}\right|=k(0-1)(1-2)(2-0)(0+1+2)$

$\implies\:1(8-2)=k(6)$

$\implies\:6=k(6)$

$\implies\:k=1$

$\implies\boxed{\bf\:\triangle=\left|\begin{array}{ccc}1&x&x^3\\1&y&y^3\\1&z&z^3\end{array}\right|=(x-y)(y-z)(z-x)(x+y+z)}$