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

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Uniteasy Admin · Accepted Answer

Using the square of a trinomial formula 


$$\because a +b+c=1$$n 
$$(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ac)$$
$$	herefore  a^2+b^2+c^2+2(ab+bc+ac) = 1 $$
$$\because a^2+b^2+c^2=2 $$n
$$ab+bc+ac = -\dfrac{1}{2}  $$n
$$\because  a^3+b^3+ c^3=3$$
$$ a^3+b^3+ c^3-3abc+3abc=3$$
$$(a+b+c)(a^2+b^2+c^2−ab−bc−ca)+3abc=3$$
$$(a+b+c)[a^2+b^2+c^2−(ab+bc+ca)] +3abc=3$$n
Plug (1), (2), (3) into (4) 
$$(2+\dfrac{1}{2})+3abc = 3   $$
Then, 
$$abc = \dfrac{1}{6}  $$n
Square the both sides of (3) 
 $$ a^2b^2+b^2c^2+a^2c^2+2abc(a+b+c) = \dfrac{1}{4}  $$
$$a^2b^2+b^2c^2+a^2c^2 = \dfrac{1}{4}-2\cdotp \dfrac{1}{6}    = - \dfrac{1}{12}  $$

$$a^4+b^4+c^4 = ( a^2+b^2+c^2)^2 - 2( a^2b^2+b^2c^2+a^2c^2)$$
$$= 4- 2	imes ( - \dfrac{1}{12}) $$
$$=4\dfrac{1}{6}  $$
Therefore, A is the choice

Multiple Choice Question (MCQ)

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