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(A) Given the quadratic equation $lx^2 + mx + n = 0$,the sum of roots is $\alpha + \beta = -m/l$ and the product of roots is $\alpha\beta = n/l$.We ne

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$l^4x^2 - nl(m^2 - 2nl)x + n^4 = 0$

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(A) Given the quadratic equation $lx^2 + mx + n = 0$,the sum of roots is $\alpha + \beta = -m/l$ and the product of roots is $\alpha\beta = n/l$.We need to find the equation with roots $S_1 = \alpha^3\beta$ and $S_2 = \alpha\beta^3$.The sum of the new roots is $S_1 + S_2 = \alpha\beta(\alpha^2 + \beta^2) = \alpha\beta((\alpha + \beta)^2 - 2\alpha\beta)$.Substituting the values: $S_1 + S_2 = (n/l)((-m/l)^2 - 2(n/l)) = (n/l)((m^2/l^2) - (2n/l)) = (n/l)((m^2 - 2nl)/l^2) = (n(m^2 - 2nl))/l^3$.The product of the new roots is $S_1 \cdot S_2 = (\alpha^3\beta)(\alpha\beta^3) = \alpha^4\beta^4 = (\alpha\beta)^4 = (n/l)^4 = n^4/l^4$.The required quadratic equation is $x^2 - (S_1 + S_2)x + (S_1 \cdot S_2) = 0$.Substituting the values: $x^2 - [n(m^2 - 2nl)/l^3]x + (n^4/l^4) = 0$.Multiplying by $l^4$,we get $l^4x^2 - nl(m^2 - 2nl)x + n^4 = 0$.

If $\alpha$ and $\beta$ are the roots of the equation $lx^2 + mx + n = 0$,then the equation whose roots are $\alpha^3\beta$ and $\alpha\beta^3$ is:

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