Question

If a, b and c are real and $ {{x}^{3}}-3{{b}^{2}}x+2{{c}^{3}} $ is divisible by $ (x-a) $ and $ (x-b), $ then

• A. $ a=-b=-c $
• B. $ a=2b=2c $
• C. $ 4{{b}^{2}}=ac $
• D. None of these

Answer 1

Answer: (D)

As $ f(x)~={{x}^{3}}-3{{b}^{2}}x+2{{c}^{2}} $ is divisible by $ (x-a) $ and $ (x-b) $ . Now, $ f(a)=0 $ $ \Rightarrow $ $ {{a}^{3}}-3{{b}^{2}}a+2{{c}^{3}}=0 $ ...(i) and $ f(b)=0 $ $ \Rightarrow $ $ {{b}^{3}}-3{{b}^{3}}+2{{c}^{3}}=0 $ ...(ii) From Eq. (ii), $ b=c $ On putting $ b=c $ in Eq. (i), we get $ {{a}^{3}}-3a{{b}^{2}}+2{{b}^{3}}=0 $ $ \Rightarrow $ $ (a-b)({{a}^{2}}+ab-2{{b}^{2}}\}=0 $ $ \Rightarrow $ $ a=b $ or $ {{a}^{2}}+ab=2{{b}^{2}} $ Thus, $ a=b=c $ or $ {{a}^{2}}+ab=2{{b}^{2}} $ and $ b=c $ Therefore, $ {{a}^{2}}+ab=2{{b}^{2}} $ and $ b=c $ is equivalent to $ a=-2b=-2c $