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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 Answer
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=?#
we know
#2(ab+bc+ca)=(a+b+c)^2-(a^2+b^2+c^2)#
#=>2(ab+bc+ca)=1^2-2=-1#
#=>ab+bc+ca=-1/2#
given
#a^3+b^3+c^3=3#
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#=>a^3+b^3+c^3-3abc+3abc=3#
#=>(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc=3#
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#=>(a+b+c)(a^2+b^2+c^2-(ab+bc+ca)+3abc=3#
#=>(1xx(2-(-1/2)+3abc))=3#
#=>(2+1/2)+3abc=3#
#=>3abc=3-5/2=1/2#
#=>abc=1/6#
Now
#(a^2b^2+b^2c^2+c^2a^2)#
#=(ab+bc+ca)^2-2ab^2c-2bc^2a-2ca^2b#
#=(ab+bc+ca)^2-2abc(b+c+a)#
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#=(-1/2)^2-2xx1/6xx1=1/4-1/3=-1/12#
Now
#a^4+b^4+c^4#
#=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)#
#=2^2-2xx(-1/12)#
#=4+1/6=4 1/6#
Extension
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#a^5+b^5+c^5#
#=(a^3+b^3+c^3)(a^2+b^2+c^2)-[a^3(b^2+c^2)+b^3(c^2+a^2)+c^3(a^2+c^2)]#
#=3*2-[a^3(b^2+c^2)+b^3(c^2+a^2)+c^3(a^2+b^2)]#
Now
#a^3(b^2+c^2)+b^3(c^2+a^2)+c^3(a^2+b^2)#
#=a^2b^2(a+b)+b^2c^2(b+c)+c^2a^2(a+c)#
#=a^2b^2(1-c)+b^2c^2(1-a)+c^2a^2(1-b)#
#=a^2b^2+b^2c^2+c^2a^2-(a^2b^2c+b^2c^2a+c^2a^2b)#
#=-1/12-abc(ab+bc+ca)#
#=-1/12-1/6*(-1/2)=0#
So
#a^5+b^5+c^5=6-0=6#
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