Using properties of determinants, show that:
$open vertical bar table row cell straight b squared straight c end cell bc cell straight b plus straight c end cell row cell straight c squared straight a squared end cell ca cell straight c plus straight a end cell row cell straight a squared straight b squared end cell ab cell straight a plus straight b end cell end table close vertical bar space equals space 0$

Solution

Let $increment space equals space open vertical bar table row cell straight b squared straight c squared end cell cell space space space bc end cell cell space straight b plus straight c end cell row cell straight c squared straight a squared end cell cell space space ca end cell cell straight c plus straight a end cell row cell straight a squared straight b squared end cell cell space ab end cell cell straight a plus straight b end cell end table close vertical bar$
$equals space straight a squared straight b squared straight c squared space open vertical bar table row bc cell space space space space 1 end cell cell space space 1 over straight b plus 1 over straight c end cell row ca cell space space space 1 end cell cell space space space space 1 over straight c plus 1 over straight a end cell row ab cell space space 1 end cell cell space space space 1 over straight a plus 1 over straight b end cell end table close vertical bar comma space by space taking space bc comma space ca comma space ab$
$common space from space straight R subscript 1 comma space straight R subscript 2 comma space straight R subscript 3$ respectively
$equals space straight a cubed straight b cubed straight c cubed space equals space space open vertical bar table row cell 1 over straight a end cell cell space space space space 1 end cell cell space space space space space 1 over straight b plus 1 over straight c end cell row cell 1 over straight b end cell cell space space space space 1 end cell cell 1 over straight c plus 1 over straight a end cell row cell 1 over straight c end cell cell space space space 1 end cell cell 1 over straight a plus 1 over straight b end cell end table close vertical bar comma space dividing space elements space of space straight C subscript 1 comma space by space abc equals space straight a cubed straight b cubed straight c cubed space equals space open vertical bar table row cell 1 over straight a end cell cell space space space space 1 end cell cell 1 over straight a plus 1 over straight b plus 1 over straight c end cell row cell 1 over straight b end cell cell space space space 1 end cell cell 1 over straight a plus 1 over straight b plus 1 over straight c end cell row cell 1 over straight c end cell cell space space 1 end cell cell 1 over straight a plus 1 over straight b plus 1 over straight c end cell end table close vertical bar comma space space by space straight C subscript 3 plus straight C subscript 1 equals space open parentheses 1 over straight a plus 1 over straight b plus 1 over straight c close parentheses straight a cubed straight b cubed straight c cubed space open vertical bar table row cell 1 over straight a end cell cell space space space space space space space 1 end cell cell space space space space space 1 end cell row cell 1 over straight b end cell cell space space space space space space 1 end cell cell space space space space space 1 end cell row cell 1 over straight c end cell cell space space space space space 1 end cell cell space space space space 1 end cell end table close vertical bar equals space straight a cubed straight b cubed straight c cubed space open parentheses 1 over straight a plus 1 over straight b plus 1 over straight c close parentheses space left parenthesis 0 right parenthesis equals space 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space two space columns space are space identical close square brackets space space space space space space space space space space space space space space space space space space space$

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