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If the zeros of the polynomial $f(x)\ =\ ax^3\ +\ 3bx^2\ +\ 3cx\ +\ d$ are in A.P., prove that $2b^3\ -\ 3abc\ +\ a^2d\ =\ 0$.
Given:
The zeros of the polynomial $f(x)\ =\ ax^3\ +\ 3bx^2\ +\ 3cx\ +\ d$ are in A.P.
To do:
Here, we have to prove that $2b^3\ -\ 3abc\ +\ a^2d\ =\ 0$.
Solution:
Let the zeros of the given polynomial be α, β and γ.
Given that the zeros are in A.P.
So, consider the roots as, $α = p – d$, $β = p$ and $γ = p +d$ where, $p$ is the first term and $d$ is the common difference.
Comparing $f(x) $ with the standard form of a cubic polynomial, $a= a$, $b= 3b$, $c= 3c$ and $d=d$
Therefore,
Sum of the roots $= α + β + γ = (p– d) + p + (p + d) = 3p = \frac{-b}{a} =\frac{-3b}{a} = -\frac{3b}{a}$
$3p= -\frac{3b}{a}$
$p= -\frac{b}{a}$
$β = p$ is a root of the given polynomial.
This implies,
$f(p)=0$
$f(-\frac{b}{a})=a(-\frac{b}{a})^3+3b(-\frac{b}{a})^2+3c(-\frac{b}{a})+d=0$
$ \begin{array}{l}
a\left(\frac{-b^{3}}{a^{3}}\right) +3b\left(\frac{b^{2}}{a^{2}}\right) +3c\left(\frac{-b}{a}\right) +d=0\\
\\
\frac{-b^{3}}{a^{2}} +\frac{3b^{3}}{a^{2}} -\frac{3bc}{a} +d=0\\
\\
\frac{-b^{3} +3b^{3} -3abc+a^{2} d}{a^{2}} =0\\
\\
2b^{3} -3abc+a^{2} d=0
\end{array}$
a\left(\frac{-b^{3}}{a^{3}}\right) +3b\left(\frac{b^{2}}{a^{2}}\right) +3c\left(\frac{-b}{a}\right) +d=0\\
\\
\frac{-b^{3}}{a^{2}} +\frac{3b^{3}}{a^{2}} -\frac{3bc}{a} +d=0\\
\\
\frac{-b^{3} +3b^{3} -3abc+a^{2} d}{a^{2}} =0\\
\\
2b^{3} -3abc+a^{2} d=0
\end{array}$
Hence proved.
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