Condition for Zeroes of a Cubic Polynomial to Be in Arithmetic Progression

Video Explanation

Question

If the zeroes of the polynomial

\[ f(x)=ax^3+3bx^2+3cx+d \]

are in arithmetic progression, prove that

\[ 2b^3-3abc+a^2d=0. \]

Solution

Step 1: Assume the Zeroes in A.P.

Let the zeroes of the polynomial be

\[ \alpha = p-q,\quad \beta = p,\quad \gamma = p+q \]

where \(p\) is the middle term and \(q\) is the common difference.

Step 2: Use Relations Between Zeroes and Coefficients

For the cubic polynomial \[ ax^3+3bx^2+3cx+d, \]

we have the standard relations:

\[ \alpha+\beta+\gamma = -\frac{3b}{a}, \]

\[ \alpha\beta+\beta\gamma+\gamma\alpha = \frac{3c}{a}, \]

\[ \alpha\beta\gamma = -\frac{d}{a}. \]

Step 3: Apply the A.P. Values

(i) Sum of the zeroes

\[ (p-q)+p+(p+q)=3p \]

So,

\[ 3p=-\frac{3b}{a} \Rightarrow p=-\frac{b}{a} \]

(ii) Sum of the products of zeroes taken two at a time

\[ (p-q)p+p(p+q)+(p-q)(p+q) \]

\[ =p^2-pq+p^2+pq+(p^2-q^2) =3p^2-q^2 \]

But,

\[ 3p^2-q^2=\frac{3c}{a} \]

Substituting \(p=-\frac{b}{a}\),

\[ 3\left(\frac{b^2}{a^2}\right)-q^2=\frac{3c}{a} \]

\[ \Rightarrow q^2=\frac{3(b^2-ac)}{a^2} \]

(iii) Product of the zeroes

\[ (p-q)p(p+q)=p(p^2-q^2) \]

But,

\[ p(p^2-q^2)=-\frac{d}{a} \]

Substituting \(p=-\frac{b}{a}\) and \(q^2=\frac{3(b^2-ac)}{a^2}\),

\[ -\frac{b}{a}\left(\frac{b^2}{a^2}-\frac{3(b^2-ac)}{a^2}\right) =-\frac{d}{a} \]

\[ -\frac{b}{a}\left(\frac{-2b^2+3ac}{a^2}\right) =-\frac{d}{a} \]

\[ \frac{2b^3-3abc}{a^3}=-\frac{d}{a} \]

Multiplying both sides by \(a^3\),

\[ 2b^3-3abc+a^2d=0 \]

Conclusion

Hence, if the zeroes of the polynomial \[ ax^3+3bx^2+3cx+d \] are in arithmetic progression, then

\[ \boxed{2b^3-3abc+a^2d=0} \]

Thus, proved.