Question

Show that $\left|\begin{array}{ccc}b+c&c+a&a+b\\a+b&b+c&c+a\\a&b&c\end{array}\right|$ = a³ + b³ + c³ - 3abc

Answer 1

Answer:

Step-by-step explanation:

to show that

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

= a³ + b³ + c³ - 3abc

split the determinant as

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

first determinant is to be zero since two rows are identical.

now solve second determinant

$\left|\begin{array}{ccc}c&a&b\\b&c&a\\a&b&c\end{array}\right|\\\\\\=c(c^{2} -ab)-a(bc-a^{2} )+b(b^{2} -ac)\\\\=c^{3} -abc-abc+a^{3}+b^{3}-abc\\\\=a^{3}+b^{3}+c^{3}-3abc$

hence proved