Question

|1 x x³ |
|1 y y³ | =(x-y)(y-z)(z-x)(x+y+z)
|1 z z ³ |
,prove it using theorems

Answer 1

Answer:

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

Step-by-step explanation:

|1 x x³ |

|1 y y³ | =(x-y)(y-z)(z-x)(x+y+z)

|1 z z ³ | 

I have applied factor theorem to solve 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

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

$\text{since }C_1\text{ and }C_2\text{ are identical, }\triangle=0$

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

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

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

$\text{Its degree is 3}$

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

$\text{Its degree is 4}$

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

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

$\implies\:\triangle=\left|\begin{array}{ccc}a1&x&x^3\\1&y&y^3\\1&z&z^3\end{array}\right|=k(x-y)(y-z)(z-x)(x+y+z)$

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

$\triangle=\left|\begin{array}{ccc}a1&0&0\\1&1&1\\1&2&8\end{array}\right|=k(0-1)(1-2)(2-0)(0+1+2)$

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

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

$\implies\:k=1$

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