Question

Prove that

|$\left[\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^3& b^3& c^3\end{array}\right]$| = $(a-b)(b-c)(c-a)(a+b+c)$

Answer 1

LHS:

$\left[\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^3& b^3& c^3\end{array}\right]$

$=1\cdot det\left(\begin{array}{cc}b& c\\ b^3& c^3\end{array}\right)-1\cdot det\left(\begin{array}{cc}a& c\\ a^3& c^3\end{array}\right)+1\cdot det\left(\begin{array}{cc}a& b\\ a^3& b^3\end{array}\right)$

$=1\cdot (b{c}^3-c{b}^3)-1\cdot (a{c}^3-c{a}^3)+1\cdot (a{b}^3-b{a}^3)$

$=b{c}^3-b^{3}c-a{c}^3+a^{3}c+a{b}^3-a^{3}b$

RHS:

$(a-b)(b-c)(c-a)(a+b+c)$

$=(ab-ac-b^2+bc)(c-a)(a+b+c)$

$=(abc-a^b-a{c}^2+a^{2}c-b^{2}c+a{b}^2+b{c}^2-abc)(a+b+c)$

$=a^2{b}^2+a{b}^3+a{b}^{2}c-a^{3}b-a^2{b}^2-a^{2}bc+a^{3}c+a^{2}bc+a^2{c}^2-a^2{c}^2-ab{c}^2-a{c}^3+ab{c}^2+b^2{c}^2+b{c}^3-a{b}^{2}c-b^{3}c-b^2{c}^2$

$=b{c}^3-b^{3}c-a{c}^3+a^{3}c+a{b}^3-a^{3}b$

LHS=RHS

Hence proved.