Question

If $\alpha ,\beta and\gamma$ are the roots of the equation $x^3$ + 2$x^2$ + 3x + 1 = 0. Find the constant term of the equation whose roots are $\frac{1}{{\beta }^3}$ + $\frac{1}{{\gamma }^3}$ - $\frac{1}{{\alpha }^3}$, $\frac{1}{{\gamma }^3}$ + $\frac{1}{{\alpha }^3}$ - $\frac{1}{{\beta }^3}$ , $\frac{1}{{\alpha }^3}$ + $\frac{1}{{\beta }^3}$ - $\frac{1}{{\gamma }^3}$.

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Answer 1

$x^3$ + 2$x^2$ + 3x + 1 = 0

$x^3$ + 1 = -(2 $x^2$ + 3x) ...................(1)

Cubing on both sides

${(x^3+1)}^3$ = - ${(2x^2+3x)}^3$

$x^9$ + 3$x^6$ + 3$x^3$ + 1 = -[8$x^6$ +27$x^3$ + 18$x^3$(2$x^2$ + 3x))]

$x^3$ + 1 = -(2$x^2$ + 3x from equation 1).

$x^9$ + 3$x^6$ + 3$x^3$ + 1 = -[8$x^6$ + 27$x^3$ + 18$x^3$(-$x^3$ -1)]

Putting $x^3$ = y in this equation

$y^3$ + 3$y^2$ +3y + 1 = -[8$y^2$ +27y + 18y (- y - 1)]

$y^3$ - 7$y^2$ +12y + 1 = 0 {its root ${\alpha }^3$, ${\beta }^3$, ${\gamma }^3$}_____________(2)

Changing y to $\frac{1}{y}$ in equation 2

$\frac{1}{y^3}$ - $\frac{7}{y^2}$ + $\frac{12}{y}$ + 1 = 0

$y^3$ + 12$y^2$ - 7y + 1 = 0 ____________(3)

If roots are {$\frac{1}{{\alpha }^3}$, $\frac{1}{{\beta }^3}$, $\frac{1}{{\gamma }^3}$}

Let $\frac{1}{{\alpha }^3}$ = a, $\frac{1}{{\beta }^3}$ = b, $\frac{1}{{\gamma }^3}$ = c

a + b + c = -12

$\frac{1}{{\beta }^3}$ + $\frac{1}{{\gamma }^3}$ - $\frac{1}{{\alpha }^3}$ = (a + b + c) - 2a = -12 - 2y

y = $\frac{12+p}{2}$

Putting the value of y in equation 3

-$\frac{{(12+p)}^3}{8}$ + 12 $\frac{{(12+p)}^2}{4}$ + 7 $\frac{(12+p)}{2}$ + 1 = 0

${(12+p)}^3$ - 24${(12+p)}^2$ - 28(12 + p) - 8 = 0

$p^3$ + 12$p^2$ - 172p - 2072 = 0

Change the variable x in place of p

$x^3$ + 12$x^2$ - 172x - 2072

Constant term = -2072