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

1.	If a+b+c=1 a^2+b^2+c^2=2 a^3+b^3+c^3=3Then find the value of , a^4+b^4+c^4.
Answer» \xa0{tex}{\\left( {a + b + c} \\right)^3} = {a^3} + {b^3} + {c^3} + \\left( {a + b + c} \\right)\\left( {ab + bc + ca} \\right){/tex}=> {tex}{\\left( 1 \\right)^3} = 3 + \\left( 1 \\right)\\left( {ab + bc + ca} \\right){/tex}=> {tex}ab + bc + ca = - 2{/tex}Squaring both sides, we get{tex}{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2} + 2abc\\left( {{a^2} + {b^2} + {c^2}} \\right) = 4{/tex}=> {tex}{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2} + 4abc = 4{/tex}=> {tex}{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2} = 4 - 4abc{/tex} ..............(i){tex}{a^2} + {b^2} + {c^2} = 2{/tex}Squaring both sides, we get=> {tex}{a^4} + {b^4} + {c^4} + 2\\left( {{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2}} \\right) = 4{/tex}=> {tex}{a^4} + {b^4} + {c^4} + 2\\left( {4 - 4abc} \\right) = 4{/tex} [From eq.(i)]=> {tex}{a^4} + {b^4} + {c^4} + 8 - 8abc = 4{/tex}=> {tex}{a^4} + {b^4} + {c^4} = - 4 + 8abc{/tex} ..................(ii)Now, using one result,\xa0{tex}6abc = {\\left( {a + b + c} \\right)^3} + 2\\left( {{a^3} + {b^3} + {c^3}} \\right) - 3\\left( {a + b + c} \\right)\\left( {{a^2} + {b^2} + {c^2}} \\right){/tex}=> {tex}6abc = {\\left( 1 \\right)^3} + 2\\left( 3 \\right) - 3\\left( 1 \\right)\\left( 2 \\right){/tex}=> {tex}abc = {1 \\over 6}{/tex}Putting this value in eq.(ii), we get{tex}{a^4} + {b^4} + {c^4} = - 4 + 8 \\times {1 \\over 6}{/tex}=> {tex}{a^4} + {b^4} + {c^4} = {{ - 8} \\over 3}{/tex}\xa0

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