Question

show that $\left[\begin{array}{ccc}b+c&c+a&a+b\\a+b&b+c&c+a\\a&b&c\end{array}\right]$ = a^3+b^3+c^3-3abc

Answer 1

Answer:

Step-by-step explanation:

We have,

$\left|\begin{array}{ccc}b+c&c+a&a+b\\a+b&b+c&c+a\\a&b&c\end{array}\right|$

$\bf{\bullet\,\,\red{R_1\rightarrow\,R_{1}+R_3}}$

$=\left|\begin{array}{ccc}a+b+c&c+a+b&a+b+c\\a+b&b+c&c+a\\a&b&c\end{array}\right|$

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

$\bf{\bullet\,\,\red{C_1\rightarrow\,C_{1}-C_2}}$

$=(a+b+c)\left|\begin{array}{ccc}0&1&1\\a-c&b+c&c+a\\a-b&b&c\end{array}\right|$

$\bf{\bullet\,\,\red{C_{2}\rightarrow\,C_{2}-C_{3}}}$

$=(a+b+c)\left|\begin{array}{ccc}0&0&1\\a-c&b-a&c+a\\a-b&b-c&c\end{array}\right|$

Expanding along $\bf{\red{R_1}}$

$=(a+b+c)\cdot1\cdot\left|\begin{array}{cc}a-c&b-a\\a-b&b-c\end{array}\right|$

$=(a+b+c)\cdot1\cdot\big\{(a-c)(b-c)-(b-a)(a-b)\big\}$

$=(a+b+c)\big\{(a-c)(b-c)+(a-b)(a-b)\big\}$

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

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

$=(a+b+c)\big(a^2+b^2+c^2-ab-bc-ca\big)$

$=a^3+b^3+c^3-3abc$