Question

# If the roots of the equation (c2−ab)x2−2(a2−bc)x+(b2−ac)=0 are real ane equal, show that either a=0 or (a3+b3+c3)=3abc.

Open in App
Solution

## Given, equation is: (c2–ab)x2–2(a2–bc)x+(b2–ac)=0To prove:a=0 or a3+b3+c3=3abc Proof: From the given equation, we have A=(c2–ab)B=–2(a2–bc)C=(b2–ac) It is being given that the equation has real and equal roots ∴D=0⇒B2–4AC=0 On substituting respective values of a, b and c in the above equation, we get [–2(a2–bc)]2–4(c2–ab)(b2–ac)=04(a2–bc)2–4(c2b2–ac3–ab3+a2bc)=04(a4+b2c2–2a2bc)–4(c2b2–ac3–ab3+a2bc)=0⇒a4+b2c2–2a2bc–b2c2+ac3+ab3–a2bc=0⇒a4+ab3+ac3–3a2bc=0⇒a[a3+b3+c3–3abc]=0⇒a=0 or a3+b3+c3=3abc

Suggest Corrections
1
Related Videos
Solving QE by Factorisation
MATHEMATICS
Watch in App