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Question

Solve the following equations :
xa+λ+yb+λ+zc+λ=1,xa+μ+yb+μ+zc+μ=1,xa+ν+yb+ν+zc+ν=1.

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Solution

Consider the following equation in t,
xa+t+yb+t+zc+t=1(tλ)(tμ)(tγ)(t+a)(t+b)(tc)
x,y,z being for the present regarded as know quantities.
This equation when cleared of fractions is of second degree in t, and is satisfied for three values t=λ,t=μ and t=v by virtue of the given equations; hence it must be an identity.To find the value of x, we multiply (1) by a+t, and then put t=a. Thus
x=(a+λ)(a+μ)(a+v)(ab)(ac)
or x=(aλ)(aμ)(av)(a+b)(a+c).
By reason of symmetry, we have
y=(b+λ)(b+μ)(b+v)(bc)(ba)
and z=(c+λ)(c+μ)(c+v)(ca)(cb)
We know that if α,β are the roots of an eqution, then this equation is
x2(α+β)x+αβ=0
or x2S1x+S2=0
where S1 is sum of roots taken one at a time and S2 is sum of products of roots taken two at a time.Similarly if there is an equation whose roots are αβγ then S1=α+β+γ,S2=αβ+βγ+γα,S3=αβγ and the corresponding equation will be t3S1t2+S2tS3=0.
Again we know that if a is a root of the equation f(x)=0 then a will satisfy the equation, i.e.f(a)=0.
Alternative Solution of Q.22 (i).
Consider the equation z+ty+t2x+t3=0
It is clear that the above equation is satisfied by t=a,b,c by virtue of the above three given equations.which is true by 1st equation and so on. Thus a,b,c are the roots of the equation.
z+ay+a2x+a3=0.
which is ture by 1st equation and so on. Thus a,b,c are the roots of the equation.
t3+t2x+ty+z=0
S1=a+b+c=x,
S2=ab+bc+ca=y,S3=abc=z
x=(a+b+c),y=ab,z=abc.

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