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If a+b+c=1,a2+b2+c2=2,a3+b3+c3=3 then find the value of a4+b4+c4=?
we know
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
a4+b4+c4
=(a2+b2+c2)2−2(a2b2+b2c2+c2a2)
=22−2×(−112)
=4+16=416
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