Question
$$\text { Show that }\left|\begin{array}{lll}x^{2} & x & 1 \\y^{2} & y & 1 \\z^{2} & z & 1\end{array}\right|=(y-z)(x-y)(x-z)$$
Step 1
We get: \[D=\left|\begin{array}{lll} x^{2} & x & 1 \\ y^{2} & y & 1 \\ z^{2} & z & 1 \end{array}\right| = x^{2}(y-z) - y^{2}(x-z) + z^{2}(x-y)\] Show more…
Show all steps
Your feedback will help us improve your experience
Taimoor Shabbir and 88 other Algebra educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Show that $\left|\begin{array}{ccc}{1} & {x} & {x^{2}} \\ {1} & {y} & {y^{2}} \\ {1} & {z} & {z^{2}}\end{array}\right|=(x-y)(y-z)(z-x)$
Systems of Equations and Inequalities
Determinants and Cramer’s Rule
Show that $$\left|\begin{array}{lll} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{array}\right|=(x-y)(y-z)(z-x)$$
Matrices and Determinants
Prove that $\left|z_{1}-z_{2}\right|^{2}=\left|z_{1}\right|^{2}-2 \operatorname{Re}\left(z_{1} \overline{z_{2}}\right)+\left|z_{2}\right|^{2} .$
Complex Numbers
The Geometry of Complex Numbers
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD