If `a_(1) x^(3) + b_(1)x^(2) + c_(1)x + d_(1) = 0 and a_(2)x^(3) + b_(2)x^(2) + c_(2)x + d_(2) = 0`
a pair of repeated roots common, then prove that
`|{:(3a_(1)", "2b_(1) ", "c_(1)),(3a_(2)", " 2b_(2)", "c_(1)),(a_(2)","b_(1)- a_(1)b_(2)", "c_(2)a_(1)-c_(2)a_(1)", "d_(1)a_(2)-d_(2)a_(1)):}|=0`

Explanation

If `f(x) = a_(1)x^(3) + b_(1) x^(2) + c_(1) x + d_(1) = ` has roots `alpha, alpha, beta,` then
`g(x) = a_(2)x^(3) + b_(2)x^(2) + c_(2)x + d_(2) = 0` must have roots `alpha, alpha, gamma`
Hence, ` a_(1) alpha^(3) + b_(1)alpha^(2)+ c_(1) alpha+ d_(1)= 0 ` (1)
` a_(2) alpha^(3) + b_(2)alpha^(2)+ c_(2) alpha+ d_(2)= 0 `(2)
Now, `alpha` is also a root of equation `f'(x) = 3a_(1)x^(2) + 2b_(1) x + c_(1) = 0` and
`g'(x) = 3 a_(2) x^(2) + 2b_(2)x + c_(2)` = 0. Therefore,
`3 a_(1) alpha^(2) + 2b_(1)alpha+ c_(1) = 0` (3)
`3 a_(2) alpha^(2) + 2b_(2)alpha+ c_(2) = 0` (4)
Also, from `a_(2) xx(1) - a_(1)xx(2)`, we have
`(a_(2)b_(1)-a_(1)b_(2) )alpha^(2) + (c_(1) a_(2) - c_(2)a_(1))alpha + d_(1)a_(2) -d_(2)a_(1)= 0` (5)
Eliminating `alpha` from (3),(4) and (5), we have
`|{:(3a_(1)", "2b_(1) ", "c_(1)),(3a_(2)", " 2b_(2)", "c_(1)),(a_(2)","b_(1)- a_(1)b_(2)", "c_(2)a_(1)-c_(2)a_(1)", "d_(1)a_(2)-d_(2)a_(1)):}|=0` .