If a + b + c = 1, ab + bc + ca = 2 and abc = 3, then the value of a⁴ + b⁴ + c⁴ is equal to ______________.

Solution

Maths sir · 4 steps

Do not hunt for a, b, c: square the sum of squares and subtract the pair terms.

1. Sum of squares. a²+b²+c² = (a+b+c)² - 2(ab+bc+ca) = 1 - 4 = -3. It is negative, so some of a, b, c are complex, but the identities still hold.
2. Squares of products. Squaring ab+bc+ca gives (ab)²+(bc)²+(ca)² + 2abc(a+b+c). So (ab)²+(bc)²+(ca)² = 2² - 2 × 3 × 1 = -2.

   a⁴+b⁴+c⁴ = (a²+b²+c²)² - 2[(ab)²+(bc)²+(ca)²]

3. Put in numbers. a⁴+b⁴+c⁴ = (-3)² - 2 × (-2) = 9 + 4 = 13.
4. So. The value of a⁴+b⁴+c⁴ is 13.

The common slip: 1. Taking (ab)²+(bc)²+(ca)² as just (ab+bc+ca)² = 4, and dropping the 2abc(a+b+c) term, gives 9 - 8 = 1, not 13.

The idea to keep. For power sums of roots, build up step by step from a+b+c, ab+bc+ca and abc. Each squaring leaves cross terms, and those cross terms are always written in the same three symmetric sums.

Answer from the NTA final answer key. Solution by PrepWiser.