# What is the sum of the squares of the roots of the cubic equation `x^3 -4x^2 +6x+1`

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Equating coefficients gives

`r_1+r_2+r_3=4` **(1)**

`r_1r_2+r_1r_3+r_2r_3=6` **(2)**

Multiply equation (1) by `r_1,` then `r_2,` then `r_3.` This gives three new equations

`r_1^2+r_1r_2+r_1r_3=4r_1`

`r_1r_2+r_2^2+r_2r_3=4r_2`

`r_1r_3+r_2r_3+r_3^2=4r_3.`

Add these to get

`r_1^2+r_2^2+r_3^2+2(r_1r_2+r_1r_3+r_2r_3)=4(r_1+r_2+r_3),`

**and using our earlier equations (1) and (2), we can solve to get**

`r_1^2+r_2^2+r_3^2=4.`

`x^3-4x^2+6x+1=0` be given equation.

Let `alpha,beta , and gamma` be the roots of above cubic equation.

By relation between roots and coefficients ,we have

`alpha+beta+gamma=-4`

`alpha beta+beta gamma+gamma alpha=6`

`alpha beta gamma=-1`

using following identity

`A^2+B^2+C^2=(A+B+C)^2-2(AB+BC+CA)`

Thus

`alpha^2+beta^2+gamma^2=(alpha+beta+gamma)^2-2(alpha beta+beta gamma+gamma alpha)`

`=(-4)^2-2(6)`

`=16-12`

`=4`

Thus our answer be

`alpha^2+beta^2+gamma^2=4`