If teh zeros of the polynomial $f (x) = a x^{3} + 3 b x^{2} + 3 c x + d$ are in A.P., prove that $2 b^{3} - 3 a b c + a^{2} d = 0$

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Solution

Let p-r, p, p+r be the zeroes of equation,

Sum of roots = p - r + p + p + r = 3p = $fraction numerator negative 3 b over denominator a end fraction$

p = $fraction numerator negative b over denominator a end fraction$

Now, p is a root of eqn. So put it in the equation:
$a left parenthesis fraction numerator negative b over denominator a end fraction right parenthesis cubed space plus 3 b left parenthesis fraction numerator negative b over denominator a end fraction right parenthesis squared minus 3 fraction numerator b c over denominator a end fraction plus d equals 0 fraction numerator negative b cubed over denominator a squared end fraction plus 3 b cubed over a squared minus 3 fraction numerator b c over denominator a end fraction plus d equals 0 minus b cubed plus 3 b cubed minus 3 a b c plus a squared d equals 0 2 b cubed minus 3 a b c plus a squared d equals 0$

Hence proved

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If zeros of f(x)= ax³+3bx²+3cx+d are in arrithmatic progression , prover that 2b³-3abc+a²d=0

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If roots of polynomial ax^3+3bx^2+3cx+d are in A.p. prove that 2b^3-3abc+a^2d=0

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