If the zeroes of the polynomial f(x) = ax³ + 3bx² + 3cx + d are in A.P., prove that 2b³ − 3abc + a²d = 0

Video Explanation

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Proof

Given polynomial:

f(x) = ax³ + 3bx² + 3cx + d

Let the zeroes of the polynomial be in A.P.

∴ Let the zeroes be: (p − q), p, (p + q)

Step 1: Use the Relationship Between Zeroes and Coefficients

For a cubic polynomial ax³ + Bx² + Cx + D:

Sum of zeroes = −B/a
Sum of products of zeroes taken two at a time = C/a
Product of zeroes = −D/a

Comparing f(x) = ax³ + 3bx² + 3cx + d with ax³ + Bx² + Cx + D:

B = 3b,   C = 3c,   D = d

Step 2: Form Equations Using A.P. Zeroes

Sum of zeroes:

(p − q) + p + (p + q) = 3p

∴ 3p = −3b/a

∴ p = −b/a

Sum of products of zeroes taken two at a time:

(p − q)p + p(p + q) + (p − q)(p + q)

= 3p² − q²

∴ 3p² − q² = 3c/a

Substituting p = −b/a:

3(b²/a²) − q² = 3c/a

∴ q² = 3(b² − ac)/a²

Step 3: Use the Product of Zeroes

Product of zeroes:

(p − q)p(p + q) = p(p² − q²)

= −d/a

Substituting p = −b/a and q² = 3(b² − ac)/a²:

(−b/a) [ (b²/a²) − 3(b² − ac)/a² ]

= −d/a

(−b/a) [ (−2b² + 3ac)/a² ] = −d/a

∴ b(2b² − 3ac) = a²d

∴ 2b³ − 3abc + a²d = 0

Hence Proved

2b³ − 3abc + a²d = 0

Conclusion

Thus, if the zeroes of the cubic polynomial f(x) = ax³ + 3bx² + 3cx + d are in arithmetic progression, then the relation 2b³ − 3abc + a²d = 0 is satisfied.