The sum of the square of the constants is 3/4
Given the equation [tex]\frac{1}{a+1} +\frac{1}{b+1} +\frac{1}{c+1} =2[/tex]
The partial form of the equation is given as:
- [tex]\frac{1}{a +1} = 2\\[/tex]
Solve for the constant "a"
[tex]\frac{1}{a +1} = 2\\a+1 = \frac{1}{2} \\a = \frac{1}{2} - 1\\a =-\frac{1}{2}[/tex]
Hence [tex]a=b=c=-\frac{1}{2} \\[/tex]
Taking the square of the constant, we will have:
[tex]a^2 + b^2 + c^2 =(-1/2)^2 +(-1/2)^2+(-1/2)^2\\a^2 + b^2 + c^2 =1/4 + 1/4 + 1/4\\a^2 + b^2 + c^2 =3/4[/tex]
Hence the sum of the square of the constants is 3/4
Learn more on partial fraction here: https://brainly.com/question/24594390
