Grade 12th passAlgebra

If 1/(a+w)+1/(b+w)+1/(c+w)+1/(d+w)=1/w where a,b,c,d belongs to real number and w is a cube root o unity then show that ∑1/(a^2-a+1)=1

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Dibyanshu

7 Years agoGrade 12th pass

1 Answer

Aditya Gupta

7 Years ago

this is a really easy ques.

1/(a+w)+1/(b+w)+1/(c+w)+1/(d+w)=1/w= w^2

we know that conjygate of w= w^2. so take conjugate on both sides and add

∑1/(a+w)+1/(a+w^2)= w^2+w= – 1

∑(2a – 1)/(a^2 – a+1)= – 1

now call ∑a/(a^2 – a+1) as y and ∑1/(a^2 – a+1) as x

so that 2y – x= – 1

or 2y+1=x

now given that ∑1/(a+w) = w^2

∑(a+w^2)/(a+w)(a+w^2) = w^2

or ∑(a+w^2)/(a^2-a+1) = w^2

or y + w^2*x= w^2

or y= w^2(1 – x)

or 2y= 2w^2(1 – x)

but 2y+1=x

so (x-1)= 2w^2(1 – x)

or (2w^2+1)(1 – x)=0

or 1-x=0

so that x = ∑1/(a^2 – a+1) = 1

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If 1/(a+w)+1/(b+w)+1/(c+w)+1/(d+w)=1/w where a,b,c,d belongs to real number and w is a cube root o unity then show that ∑1/(a^2-a+1)=1

Dibyanshu

1 Answer

Aditya Gupta

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