﻿ If a+b+c=1, a^2+b^2+c^2=2,a^3+b^3+ c^3=3, then find the value of a^4+b^4+c^4

Multiple Choice Question (MCQ)

If a+b+c=1, a^2+b^2+c^2=2,a^3+b^3+ c^3=3, then find the value of a^4+b^4+c^4

1. 4\dfrac{1}{6}

2. ×

4\dfrac{5}{6}

3. ×

4\dfrac{2}{3}

4. ×

4

Collected in the board: Calculate algebraic expressions

Steven Zheng posted 2 years ago

1. Using the square of a trinomial formula

\because a +b+c=1
(1)

(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ac)

\therefore a^2+b^2+c^2+2(ab+bc+ac) = 1

\because a^2+b^2+c^2=2
(2)
ab+bc+ac = -\dfrac{1}{2}
(3)
\because a^3+b^3+ c^3=3

a^3+b^3+ c^3-3abc+3abc=3

(a+b+c)(a^2+b^2+c^2−ab−bc−ca)+3abc=3

(a+b+c)[a^2+b^2+c^2−(ab+bc+ca)] +3abc=3
(4)

Plug (1), (2), (3) into (4)

(2+\dfrac{1}{2})+3abc = 3

Then,

abc = \dfrac{1}{6}
(5)

Square the both sides of (3)

a^2b^2+b^2c^2+a^2c^2+2abc(a+b+c) = \dfrac{1}{4}

a^2b^2+b^2c^2+a^2c^2 = \dfrac{1}{4}-2\cdotp \dfrac{1}{6} = - \dfrac{1}{12}

a^4+b^4+c^4 = ( a^2+b^2+c^2)^2 - 2( a^2b^2+b^2c^2+a^2c^2)

= 4- 2\times ( - \dfrac{1}{12})

=4\dfrac{1}{6}

Therefore, A is the choice

Steven Zheng posted 2 years ago

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