With three letters a,b,c the examiner gives you a couple of symmetric totals — usuallya+b+c and ab+bc+ca — and asks for something bigger. Two formulas build everything from those two givens.
Two engines for three variables
• The master square:(a+b+c)2=a2+b2+c2+2(ab+bc+ca), so a2+b2+c2=(a+b+c)2−2(ab+bc+ca).
• The cube factor:a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca).
• Golden special case: if a+b+c=0 then a3+b3+c3=3abc. This single fact answers a huge share of three-variable questions.
A handy spin-off: (a−b)2+(b−c)2+(c−a)2=2[(a2+b2+c2)−(ab+bc+ca)].
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=6 and ab+bc+ca=11, then a2+b2+c2=36−22= .
The same square-then-square idea powers the higher-degree ladders: to getx4+x41 you square x2+x21 and subtract 2; to climb down, add 2 and take the square root.
📝Practice Questions
Q1If a + b + c = 7 and ab + bc + ca = 14, find a² + b² + c².
Q2If x + y + z = 0, then x³ + y³ + z³ equals:
Q3If x + 1/x = 3, find x⁴ + 1/x⁴.
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Real exam questions — Three-Variable & Higher Powers
2 question types · 8 solved examples from real SSC papers
Three-letter symmetric sums and high powers all run off two engines: the square of a sum and the cube factor identity. Build the missing piece step by step.
How to solve this type
The master square is (a+b+c)2=a2+b2+c2+2(ab+bc+ca), so a2+b2+c2=(a+b+c)2−2(ab+bc+ca). The big factor identity is a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca). A golden special case: if a+b+c=0, then a3+b3+c3=3abc exactly.
If x+y+z=8 and xy+yz+zx=14, find x2+y2+z2.
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If a+b+c=5 and ab+bc+ca=7, find (a−b)2+(b−c)2+(c−a)2.
