Question

If the roots of the equation \[ x^3+3px^2+3qx+r=0 \] are in geometric progression, then:

• A. \(p^3r+q^3=0\)
• B. \(p^3r-q^3=0\)
• C. \(p^3+rq^2=0\)
• D. \(p^3-rq^2=0\)

Hint:

For cubic roots in G.P., assume roots as \(\frac{a}{k},a,ak\) and apply Vieta's formula.

Answer 1

The Correct Option is B

Concept:
For a cubic equation, relation between roots and coefficients is obtained using Vieta's formula.

Step 1: Let roots be in G.P.
Let the roots be: \[ \frac{a}{k},\quad a,\quad ak \]

Step 2: Use product of roots.
For: \[ x^3+3px^2+3qx+r=0 \] Product of roots: \[ \frac{a}{k}\cdot a\cdot ak=a^3=-r \] \[ a^3=-r \]

Step 3: Use sum and pairwise sum.
Sum of roots: \[ a\left(\frac{1}{k}+1+k\right)=-3p \] Pairwise sum: \[ a^2\left(\frac{1}{k}+1+k\right)=3q \] Dividing pairwise sum by sum: \[ \frac{3q}{-3p}=a \] \[ a=-\frac{q}{p} \]

Step 4: Substitute in product relation. \[ a^3=-r \] \[ \left(-\frac{q}{p}\right)^3=-r \] \[ -\frac{q^3}{p^3}=-r \] \[ p^3r=q^3 \] \[ p^3r-q^3=0 \] \[ \therefore \text{Correct Answer is (B)} \]