These questions hand you a few facts about three numbers a, b, c — their sum, their product, the sum of their squares — and ask for something else. Two identities connect everything.
The two identities — and one magic shortcut
Square rule: (a+b+c)² = a²+b²+c² + 2(ab+bc+ca). This links the sum, the sum of squares, and the pairwise products — given any two, find the third.
Golden shortcut: if a+b+c = 0, the bracket kills the right side, so a³+b³+c³ = 3abc. a+b+c = 6 ≠ 0
a²+b²+c² = 14 a³+b³+c³ = 36
If a + b + c = 0, then a³ + b³ + c³ equals .
📝Practice Questions
Q1If a + b + c = 0, then a³ + b³ + c³ equals:
Q2If a+b+c = 6 and a²+b²+c² = 14, then ab+bc+ca is:
Q3For a=2, b=3, c=4, the value of a³+b³+c³ − 3abc is:
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Real exam questions — Three-Variable Identities
2 question types · 6 solved examples from real SSC papers
Two identities run this topic — the square rule and the cube rule — plus the magic a+b+c=0 shortcut.
How to solve this type
Two identities run this whole topic. First, the square rule: (a+b+c)² = a²+b²+c² + 2(ab+bc+ca). Second, the cube rule: a³+b³+c³ − 3abc = (a+b+c)(a²+b²+c² − ab−bc−ca). The single most useful fact follows from the cube rule: if a+b+c = 0, the bracket (a+b+c) makes the whole right side 0, so a³+b³+c³ = 3abc. Spot a+b+c = 0 first and you often skip all the heavy arithmetic.
If a+b+c=0, what is abca3+b3+c3?
A1B3C6D0
If a=1, b=2, c=3, find a3+b3+c3−3abc:
A12B18C6D0
If a+b+c=12, a2+b2+c2=50 and abc=60, find a3+b3+c3:
