QJEEM21Mafd5a: JEE Main 2021 Mathematics - Class 11 JEEM21Mafd5a Solution: If a + b + c = 1, ab + bc + ca = 2 and abc = 3, then the value…

(a + b + c)² = 1

$$\Rightarrow$$ a² + b² + c² + 2(ab + bc + ca) = 1

$$\Rightarrow$$ a² + b² + c² = – 3 ….(i)

$$\Rightarrow$$ ab + bc + ca = 2 ….(ii)

Squaring of equation (ii),

$$\Rightarrow$$ a²b² + b²c² + c²a² + 2(ab²c + bc²a + ca²b) = 4

$$\Rightarrow$$ a²b² + b²c² + c²a² + 2abc(a + b + c) = 4

$$\Rightarrow$$ a²b² + b²c² + c²a² + 6 = 4

$$\Rightarrow$$ a²b² + b²c² + c²a² = – 2 ….(iii)

Squaring of equation (i),

$$\Rightarrow$$ a⁴ + b⁴ + c⁴ + 2(a²b² + b²c² + c²a²) = 9

$$\Rightarrow$$ a⁴ + b⁴ + c⁴ – 4 = 9

$$\Rightarrow$$ a⁴ + b⁴ + c⁴ = 13