[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) Reply
Answer:
a3+b3+c3-3×1
Step-by-step explanation:
a3+b3+c3-3
[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: