Answers

The Brainliest Answer!
2014-04-19T16:50:02+05:30
 x^{2} (px+q)-rx-1=0
px^3+qx^2-rx-r=0
as for cubic equation
 abc= \frac{-r}{p}
a+b+c=\frac{-q}{p}
ab+bc+ca= \frac{r}{p}
now the value of determinant is
=(1+a)(1+b)(1+c)-1-a-b-c=
=ab+bc+ca+abc
as
 abc= \frac{-r}{p}
ab+bc+ca= \frac{r}{p}
ab+bc+ca+abc= \frac{r}{p}- \frac{r}{p}=0
So the Value of determinant is zero

2 4 2
  • Brainly User
2014-04-19T17:33:34+05:30
The value of determinat is abc(1+1/a+1/b+1/c)
found using elementary operations
and form the cubic equation u can say that it is 0
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