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  1. If (2a – 3)2 + (3b + 4)2 + (6c + 1)2 = 0, then the value of
    a3 + b3 + c3 − 3ac
    + 3 is :
    a2 + b2 + c2
    1. abc + 3
    2. 6
    3. 0
    4. 3
Correct Option: D

(2a – 3)2 + (3b + 4)2 + (6c + 1)2 = 0

∴  2a – 3 = 0 ⇒ a =
3
,
2

3b + 4 = 0 ⇒ b =
−4
3

6c + 1 = 0 ⇒ c = –
1
6

∴ a + b + c =
3
4
1
236

=
9 − 8 − 1
= 0
6

∴  a3 + b3 + c3 − 3ac = 0
∴ 
a3 + b3 + c3 − 3ac
+ 3
a2 + b2 + c2

= 0 + 3 = 3



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