Question

a+b+c=2 a^2+b^2+c^2=6 a^3+b^3+c^3=8 a^4+b^4+c^4=​

Answer 1

Answer: 2(ab+bc+ca)=(a+b+c)2−(a2+b2+c2)

⇒2(ab+bc+ca)=12−2=−1

⇒ab+bc+ca=−12

given

a3+b3+c3=3

⇒a3+b3+c3−3abc+3abc=3

⇒(a+b+c)(a2+b2+c2−ab−bc−ca)+3abc=3

⇒(a+b+c)(a2+b2+c2−(ab+bc+ca)+3abc=3

⇒(1×(2−(−12)+3abc))=3

⇒(2+12)+3abc=3

⇒3abc=3−52=12

⇒abc=16

Now

(a2b2+b2c2+c2a2)

=(ab+bc+ca)2−2ab2c−2bc2a−2ca2b

=(ab+bc+ca)2−2abc(b+c+a)

=(−12)2−2×16×1=14−13=−112

Now

a4+b4+c4

=(a2+b2+c2)2−2(a2b2+b2c2+c2a2)

=22−2×(−112)

=4+16=416

Extension

a5+b5+c5

=(a3+b3+c3)(a2+b2+c2)−[a3(b2+c2)+b3(c2+a2)+c3(a2+c2)]

=3⋅2−[a3(b2+c2)+b3(c2+a2)+c3(a2+b2)]

Now

a3(b2+c2)+b3(c2+a2)+c3(a2+b2)

=a2b2(a+b)+b2c2(b+c)+c2a2(a+c)

=a2b2(1−c)+b2c2(1−a)+c2a2(1−b)

=a2b2+b2c2+c2a2−(a2b2c+b2c2a+c2a2b)

=−112−abc(ab+bc+ca)

=−112−16⋅(−12)=0

So

a5+b5+c5=6−0=6

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