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If alpha, beta, gamma are the roots of the cubic equation x^(3)+qx+r=0 then the find equation whose roots are (alpha-beta)^(2),(beta-gamma)^(2),(gamma-alpha)^(2). |
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Answer» Solution :`:' alpha, beta, gamma` are the roots of the cubic EQUATION `x^(3)+qx+r=0`………i Then `SUM alpha=0, sum alpha beta=q, alpha beta gamma =-r`……..ii If y is a root of the REQUIRED equation, then `y=(alpha-beta)^(2)=(alpha+beta)^(2)-4 alpha beta` `=(alpha+beta+gamma-gamma)^(2)-(4 alpha beta gamma)/(gamma)` `=(0-gamma)^(2)+(4r)/(gamma)` [from Eq. (ii) ]` `impliesy=gamma^(2)+(4r)/(gamma)` [replacing `gamma` by `x` which is a root of Eq. (i) ] `:.y=x^(2)+(4r)/x` Or `x^(3)-yx+4r=0`..........iii The required equation is obatained by eliminating `x` between Eqs (i) and (iii) Now, subtracting Eq. (iii) from Eq. (i) we get `(q+y)x-3r=0` or `x=(3r)/(q+y)` On substituting the value of `x` in Eq. (i) we get `((3r)/(q+y))^(3)+q((3r)/(q+y))+r=0` Thus, `y^(3)+6qy^(2)+9q^(2)y+(4q^(3)+27r^(2))=0` which is the required equation. |
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