if a+ b +c= 1 and a2 +b2 +c2= 83 find value of a3 + b3 +c3 -3abc

if a+ b +c= 1 and a2 +b2 +c2= 83 find value of a3 + b3 +c3 -3abc

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2 thoughts on “if a+ b +c= 1 and a2 +b2 +c2= 83 find value of a3 + b3 +c3 -3abc”

  1. [tex]\begin{gathered}\begin{gathered}\bf \:Given – \begin{cases} &\sf{a + b + c = 1} \\ &\sf{ {a}^{2} + {b}^{2} + {c}^{2} = 83 } \end{cases}\end{gathered}\end{gathered}[/tex]

    [tex]\begin{gathered}\begin{gathered}\bf \: To\: find – \begin{cases} &\sf{ {a}^{3} + {b}^{3} + {c}^{3} – 3abc }  \end{cases}\end{gathered}\end{gathered}[/tex]

    [tex]\begin{gathered}\Large{\sf{{\underline{Formula \: Used – }}}}  \end{gathered}[/tex]

    [tex]1. \: \sf \: {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca[/tex]

    [tex]2. \: \sf \: {a}^{3} + {b}^{3} + {c}^{3} – 3abc= (a + b + c)( {a}^{2} + {b}^{2} {c}^{2} – ab – bc – ca) [/tex]

    [tex]\large\underline{\sf{Solution-}}[/tex]

    [tex] \rm :\longmapsto\:\sf \: a \: + \: b \: + \: c \: = \: 1[/tex]

    On squaring both sides, we get

    [tex]\rm :\longmapsto\: \sf \: {(a + b + c)}^{2} = 1[/tex]

    [tex]\rm :\longmapsto\: \sf \: {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca = 1[/tex]

    [tex] \rm :\longmapsto\:\sf \: 83 + 2ab + 2bc + 2ca = 1[/tex]

    [tex] \rm :\longmapsto\:\sf \: 2ab + 2bc + 2ca = 1 – 83[/tex]

    [tex] \rm :\longmapsto\:\sf \: 2ab + 2bc + 2ca = – 82[/tex]

    [tex]\rm :\longmapsto\: \sf \: ab + bc + ca = – \: 41[/tex]

    Now,

    [tex]\rm :\longmapsto\:\sf \: {a}^{3} + {b}^{3} + {c}^{3} – 3abc[/tex]

    [tex] \sf \: = \: (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} – ab – bc – ca) [/tex]

    [tex] \sf \: = \: (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} – (ab + bc + ca) ) [/tex]

    [tex] \sf \: = \: (1)\bigg(83 – ( – 41) \bigg) [/tex]

    [tex] \sf \: = \: 83 + 41[/tex]

    [tex] \sf \: = \: 124[/tex]

    More Identities to know:

    • (a + b)² = a² + 2ab + b²
    • (a – b)² = a² – 2ab + b²
    • a² – b² = (a + b)(a – b)
    • (a + b)² = (a – b)² + 4ab
    • (a – b)² = (a + b)² – 4ab
    • (a + b)² + (a – b)² = 2(a² + b²)
    • (a + b)³ = a³ + b³ + 3ab(a + b)
    • (a – b)³ = a³ – b³ – 3ab(a – b)
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