Shortcut Trick
Direct formula for a3 + b3 + c3 − 3abc using sum and pairwise product sum:
Value = (a + b + c) × [(a + b + c)2 − 3(ab + bc + ca)]
Substitute values: 6 × [62 − 3 × 4]
⇒ 6 × [36 − 12] = 6 × 24 = 144
∴ The correct answer is 144.
Alternate Method
Given: a + b + c = 6 and ab + bc + ca = 4
Method: Algebraic Identity expansion
Step 1: Find the value of a2 + b2 + c2 using the square of a trinomial.
⇒ (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
⇒ 62 = a2 + b2 + c2 + 2(4)
⇒ 36 = a2 + b2 + c2 + 8
⇒ a2 + b2 + c2 = 36 − 8 = 28
Step 2: Use the standard identity for a3 + b3 + c3 − 3abc.
⇒ a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − (ab + bc + ca))
⇒ a3 + b3 + c3 − 3abc = 6 × (28 − 4)
⇒ a3 + b3 + c3 − 3abc = 6 × 24 = 144
∴ The correct answer is 144.
Additional Information
Zero Sum Condition
If a + b + c = 0, then a3 + b3 + c3 = 3abc. This is a very common variation in exams.
Alternative Form of Identity
a3 + b3 + c3 − 3abc = 0.5 × (a + b + c)[(a − b)2 + (b − c)2 + (c − a)2], used when differences are given.
Sum of Squares Identity
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) is the fundamental relation used to link these variables.